summary：44

沒救了。。。整天刷水遲早藥丸！

?bzoj3892：

區間dp。我原來的思路是dp[i][j]表示前i個數跳過了j次，那么轉移可以前k個數轉移了j-1次，枚舉k就好了，但是這樣是錯的，因為前k個數轉移了j-1次，那么再k到i之間到底要在哪一步跳無法確定。于是便WA了。正確的轉移應該是前i-k-1個數轉移了j-k次，在i-k-1到i直接跳過去就好了。dp的轉移方程要多驗證是否是對的。。。！

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}while(isdigit(c)) x=x*10+c-'0',c=getchar();return x*f; } const int nmax=505; const int inf=0x7f7f7f7f; int dp[nmax][nmax],x[nmax],y[nmax]; int as(int x){return x<0?-x:x; } int main(){int N=read(),K=read();rep(i,1,N) x[i]=read(),y[i]=read();clr(dp,0x7f);dp[1][0]=0;rep(i,2,N) {for(int j=0;j<=i-2&&j<=K;j++){rep(k,0,j)dp[i][j]=min(dp[i][j],dp[i-k-1][j-k]+as(x[i]-x[i-k-1])+as(y[i]-y[i-k-1]));}}printf("%d\n",dp[N][min(K,N-2)]); }

bzoj3891：

直接三次bfs就好了。我傻叉寫了三個bfs。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=40005; const int maxn=100005; const int inf=0x7f7f7f7f; struct edge{int to;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v){pt->to=v;pt->next=head[u];head[u]=pt++;pt->to=u;pt->next=head[v];head[v]=pt++; } int dist[nmax],dis[nmax],d[nmax],B,E,P,N,M; bool vis[nmax]; queue<int>q; void d1(){clr(vis,0);vis[1]=1;d[1]=0;q.push(1);while(!q.empty()){int x=q.front();q.pop();qwq(x) if(!vis[o->to]){d[o->to]=d[x]+1;q.push(o->to);vis[o->to]=1;}} } void d2(){clr(vis,0);vis[2]=1;dis[2]=0;q.push(2);while(!q.empty()){int x=q.front();q.pop();qwq(x) if(!vis[o->to]){dis[o->to]=dis[x]+1;q.push(o->to);vis[o->to]=1;}} } void d3(){clr(vis,0);vis[N]=1;dist[N]=0;q.push(N);while(!q.empty()){int x=q.front();q.pop();qwq(x) if(!vis[o->to]){dist[o->to]=dist[x]+1;q.push(o->to);vis[o->to]=1;}} } int main(){B=read(),E=read(),P=read(),N=read(),M=read();int u,v,tmp,temp;rep(i,1,M) u=read(),v=read(),add(u,v);d1();d2();d3();int ans=inf;rep(i,1,N) ans=min(ans,B*d[i]+E*dis[i]+P*dist[i]);printf("%d\n",ans);return 0; }

bzoj3943：

最大生成樹。prim O(n2) kruskal O(eloge) 這道題邊數很多用prim快了很多。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) #define ll long long int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=2005; const int maxn=4000005; const int inf=0x7f7f7f7f; int a[nmax],n,f[nmax];ll ans; bool vis[nmax]; int main(){n=read();rep(i,1,n) a[i]=read();int tmp;rep(i,1,n){tmp=0;rep(j,1,n) if(!vis[j]&&f[j]>=f[tmp]) tmp=j;vis[tmp]=true;ans+=f[tmp];rep(j,1,n) f[j]=max(f[j],a[tmp]^a[j]);}printf("%lld\n",ans);return 0; } /*struct node{int from,to,dist;bool operator<(const node&rhs)const{return dist>rhs.dist;} }; node nodes[maxn]; int a[nmax],fa[nmax]; int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]); } int main(){int n=read(),cnt=0;rep(i,1,n) {a[i]=read();rep(j,1,i-1){nodes[++cnt].from=j,nodes[cnt].to=i,nodes[cnt].dist=a[i]^a[j];}}sort(nodes+1,nodes+cnt+1);rep(i,1,n) fa[i]=i;int ta,tb,sum=0;ll ans=0;rep(i,1,cnt){ta=find(nodes[i].from);tb=find(nodes[i].to);if(ta!=tb){fa[ta]=tb;sum++;ans+=nodes[i].dist;if(sum==n-1) break;}}printf("%lld\n",ans);return 0; }*/

bzoj3942：

AC自動機強上。。判斷是否是子串就可以了。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=1000005; int ch[nmax][26],fail[nmax],F[nmax],T[nmax]; char s[nmax],t[nmax],S[nmax]; void insert(){int len=strlen(s),t=0,pt=0;rep(i,0,len-1){if(!ch[t][s[i]-'a']) ch[t][s[i]-'a']=++pt;t=ch[t][s[i]-'a'];}F[t]=len; } queue<int>q; void getfail(){fail[0]=0;q.push(0);while(!q.empty()){int x=q.front();q.pop();rep(i,0,25) {if(ch[x][i]) q.push(ch[x][i]),fail[ch[x][i]]=x==0?0:ch[fail[x]][i];else ch[x][i]=x==0?0:ch[fail[x]][i];}} } int main(){scanf("%s",t+1);scanf("%s",s);insert();getfail();int len=strlen(t+1),top=0,x=0;rep(i,1,len){S[++top]=t[i];x=ch[x][t[i]-'a'];T[top]=x;if(F[x]) top-=F[x],x=T[top];}rep(i,1,top) printf("%c",S[i]);printf("\n");return 0; }

bzoj3412：

lower_bound查找即可。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=50005; int sum[nmax]; int main(){int n=read(),Q=read(),u,tmp,temp;sum[0]=-1;rep(i,1,n) u=read(),sum[i]=sum[i-1]+u;rep(i,1,Q){u=read();tmp=lower_bound(sum+1,sum+n+1,u)-sum;printf("%d\n",tmp);}return 0; }

bzoj3431：

在枚舉距離的同時枚舉時間。不錯的模擬！

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=10005; double a[nmax],b[nmax]; char s[5]; int main(){int n,asum=0,bsum=0;scanf("%d",&n);double u,tmp,temp;rep(i,1,n){scanf("%s%lf",s,&u);if(s[0]=='T') a[++asum]=u;else b[++bsum]=u;}b[++bsum]=1000;sort(a+1,a+asum+1);sort(b+1,b+bsum+1);int v=1,now=1;double pre=0,dist;rep(i,1,bsum){dist=b[i]-b[i-1];while(pre+dist*v>a[now]&&now<=asum){dist-=(a[now]-pre)/v;pre=a[now];now++;v++;}pre+=dist*v;v++;//printf("%lf %d\n",pre,v);}int ans=pre+0.5;printf("%d\n",ans);return 0; }

bzoj3432:

二分答案即可。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();int f=1;while(!isdigit(c)) {if(c=='-') f=-1;c=getchar();}while(isdigit(c)) x=x*10+c-'0',c=getchar();return x*f; } const int nmax=505; const int inf=0x7f7f7f7f; int map[nmax][nmax],t[nmax][nmax]; bool vis[nmax][nmax]; int N,M,cnt=0,ta,tb;struct node{int x,y;node(int x,int y):x(x),y(y){}; }; queue<node>q; int as(int x){return x<0?-x:x; }int xx[5]={0,0,0,1,-1}; int yy[5]={0,1,-1,0,0}; bool check(int mid){int x,y,tx,ty,res=0;clr(vis,0);vis[ta][tb]=1;q.push(node(ta,tb));while(!q.empty()){node o=q.front();q.pop();x=o.x,y=o.y;if(t[x][y]==1) res++;rep(i,1,4){tx=x+xx[i],ty=y+yy[i];if(tx<1||ty<1||tx>N||ty>M||as(map[tx][ty]-map[x][y])>mid||vis[tx][ty]) continue;q.push(node(tx,ty));vis[tx][ty]=1;}}return res==cnt; } int main(){N=read(),M=read();rep(i,1,N) rep(j,1,M) map[i][j]=read();rep(i,1,N) rep(j,1,M) {t[i][j]=read();if(t[i][j]==1) cnt++,ta=i,tb=j;}int l=0,r=inf,mid,ans=0;while(l<=r){mid=(l+r)>>1;if(check(mid)) ans=mid,r=mid-1;else l=mid+1;}printf("%d\n",ans);return 0; }

bzoj3410：

求最多能用多少個不相交的區間覆蓋線段。貪心：按結束點排序，放不了就不要。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=50005; int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } struct node{int l,r;bool operator<(const node&rhs)const{return r<rhs.r;} }; node nodes[nmax]; int main(){int N=read(),u,v;rep(i,1,N) nodes[i].l=read(),nodes[i].r=read();sort(nodes+1,nodes+N+1);int ans=0,now=-1;rep(i,1,N) {if(nodes[i].l>=now) ans++,now=nodes[i].r;}printf("%d\n",ans);return 0; }

?bzoj3404：

對于游戲問題不怎么懂。。?？梢韵阮A處理出每個數是必勝/必敗，該狀態是必勝的當后續狀態有狀態是必敗的，否則就是必敗的。因為在預處理的時候第一次出現必敗的情況他的后續狀態都是必勝的，所以每個數都只有必勝或必敗兩種情況。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=1000005; int ans[nmax]; int main(){clr(ans,0);rep(i,1,9) ans[i]=1;int maxn,minn,tmp,temp;rep(i,10,nmax){maxn=-1,minn=10;tmp=i;while(tmp){temp=tmp%10;if(temp){maxn=max(maxn,temp);minn=min(minn,temp);}tmp/=10;}if(maxn!=-1) ans[i]|=(!ans[i-maxn]);if(minn!=10) ans[i]|=(!ans[i-minn]);}int G;scanf("%d",&G);rep(i,1,G){scanf("%d",&tmp);if(ans[tmp]) printf("YES\n");else printf("NO\n");}return 0; }

bzoj3433：

貪心。按結束點排序，優先進入較長的那個隊。好神的貪心！

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=155; struct node{int l,r;bool operator<(const node&rhs)const{return r<rhs.r;} }; node e[nmax]; int main(){int n;scanf("%d",&n);rep(i,1,n){scanf("%d%d",&e[i].l,&e[i].r);e[i].r--;}sort(e+1,e+n+1);int now=-1,pre=-1,ans=0;rep(i,1,n){if(e[i].l>now) ans++,now=e[i].r;else if(e[i].l>pre) ans++,pre=e[i].r;if(pre>now) swap(now,pre);}printf("%d\n",ans);return 0; }

bzoj3403：

模擬。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) const int nmax=200005; int a[nmax]; char s[5],ch[5]; int main(){int N,l=100001,r=100000,cnt=0,tmp;scanf("%d",&N);rep(i,1,N){scanf("%s%s",s,ch);if(s[0]=='A'){cnt++;if(ch[0]=='L') a[--l]=cnt;else a[++r]=cnt;}else{scanf("%d",&tmp);if(ch[0]=='L') l+=tmp;else r-=tmp;}}rep(i,l,r) printf("%d\n",a[i]);return 0; }

bzoj3402：

邊權為1，我傻叉還打了個dijkstra。。應該bfs一遍就好了吧。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define qwq(x) for(edge *o=head[x];o;o=o->next) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=20005; const int maxn=100005; const int inf=0x7f7f7f7f;struct edge{int to;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v){pt->to=v;pt->next=head[u];head[u]=pt++;pt->to=u;pt->next=head[v];head[v]=pt++; }struct node{int x,dist;node(int x,int dist):x(x),dist(dist){};bool operator<(const node&rhs)const{return dist>rhs.dist;} }; priority_queue<node>q; int dist[nmax]; void dijkstra(int s){q.push(node(s,0));clr(dist,0x7f);dist[s]=0;int x,d;while(!q.empty()){node o=q.top();q.pop();x=o.x;d=o.dist;if(d!=dist[x]) continue;qwq(x) if(dist[o->to]>d+1){dist[o->to]=d+1;q.push(node(o->to,dist[o->to]));}} } int main(){int N=read(),M=read(),u,v;rep(i,1,M){u=read(),v=read();add(u,v);}dijkstra(1);int ans=0;rep(i,1,N) ans=max(ans,dist[i]);int res,cnt=0;dwn(i,N,1) if(dist[i]==ans) res=i,cnt++;printf("%d %d %d\n",res,ans,cnt);return 0; }

bzoj3401：

又是單調棧。。。usaco相似題怎么那么多

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=1e5+5; int a[nmax],s[nmax],ans[nmax]; int main(){int N=read(),r=0;rep(i,1,N){a[i]=read();while(r&&a[s[r]]<a[i]) ans[s[r]]=i,r--;s[++r]=i;}rep(i,1,N) printf("%d\n",ans[i]);return 0; }

bzoj3400：

dp?？梢詨旱粢痪S。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=1005; const int maxn=2005; const int mod=100000000; int dp[maxn][nmax]; int main(){int N=read(),F=read(),u;u=read();dp[1][u%F]=1;dp[1][0]+=1;rep(i,2,N){u=read();rep(j,0,F-1) {dp[i][j]=(dp[i-1][j]+dp[i][j])%mod;dp[i][(j+u)%F]=(dp[i-1][j]+dp[i][(j+u)%F])%mod;}}printf("%d\n",dp[N][0]-1);return 0; }//其實可以壓掉一維。。如下。 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=1005; const int mod=100000000; int dp[nmax]; int pre[nmax]; int main(){int N=read(),F=read(),u,v;u=read();pre[u%F]=1;pre[0]++;rep(i,2,N){u=read();rep(j,0,F-1) {dp[j]=(dp[j]+pre[j])%mod;v=(j+u)%F;dp[v]=(pre[j]+dp[v])%mod;}rep(j,0,F-1) pre[j]=dp[j],dp[j]=0;}printf("%d\n",pre[0]-1);return 0; }

bzoj3392：

bfs。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=1005; const int maxn=50005; const int inf=0x7f7f7f7f;struct edge{int to;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v){pt->to=v;pt->next=head[u];head[u]=pt++; }int q[nmax],dist[nmax]; int main(){int N=read(),T=read(),u,v;rep(i,1,N) {u=read(),v=read();add(u,v);}int l=1,r=1;q[1]=1;while(l<=r){u=q[l];l++;qwq(u) if(!dist[o->to]){dist[o->to]=dist[u]+1;q[++r]=o->to;if(o->to==T) {printf("%d\n",dist[o->to]+1);return 0;}}}printf("-1\n");return 0; }

bzoj3016：

括號序列！以前看過然后不會。。。貪心！！！

#include<cstdio> int main(){char c;int ans=0,cnt=0;while(1){c=getchar();if(c=='\n') break;if(c=='(') cnt++;else if(c==')'){cnt--;if(cnt<0) cnt+=2,ans++;}}printf("%d\n",ans+(cnt>>1));return 0; }

bzoj2019：

直接將點的權值裝換到邊上，然后判斷是否有正環（-），spfa就好了。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<deque> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=225; const int maxn=1005; const int inf=0x7f7f7f7f;struct edge{int to,dist;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v,int d){pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++; }queue<int>q; int dist[nmax],in[nmax]; bool inq[nmax]; bool spfa(int s,int D,int N){clr(dist,0);dist[s]=-D;clr(inq,0);in[s]=1;q.push(s);while(!q.empty()){int x=q.front();q.pop();inq[x]=0;qwq(x) if(dist[o->to]>dist[x]+o->dist){dist[o->to]=dist[x]+o->dist;if(!inq[o->to]){in[o->to]++;if(in[o->to]>N) return 0;q.push(o->to);inq[o->to]=1;}}}return 1; }int main(){int D=read(),M=read(),N=read(),P=read(),S=read(),u,v,d;rep(i,1,M) {u=read();v=read(),add(u,v,-D);}rep(i,1,P) {u=read();v=read();d=read();add(u,v,d-D);}if(spfa(S,D,N)) {int ans=0;rep(i,1,N) ans=max(ans,-dist[i]);printf("%d\n",ans);}else printf("-1\n");return 0; }

?bzoj2101：

神dp。。。然后可以枚舉間距巧妙壓掉一維。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=5005; int f[nmax],a[nmax],sum[nmax]; int main(){int N=read();rep(i,1,N) a[i]=read(),sum[i]=sum[i-1]+a[i];rep(i,0,N-1) rep(j,1,N-i) {f[j]=max(sum[j+i]-sum[j-1]-f[j],sum[j+i]-sum[j-1]-f[j+1]);}printf("%d\n",f[1]);return 0; }

bzoj2100：

兩遍dijkstra就好了。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) const int nmax=100005; const int maxn=400005; const int inf=0x7f7f7f7f; int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } struct edge{int to,dist;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v,int d){pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++; } struct node{int x,dist;node(int x,int dist):x(x),dist(dist){};bool operator<(const node&rhs)const{return dist>rhs.dist;} }; priority_queue<node>q; int dist[nmax]; void dijkstra(int s){clr(dist,0x7f);dist[s]=0;q.push(node(s,0));while(!q.empty()){node o=q.top();q.pop();int x=o.x,d=o.dist;if(dist[x]!=d) continue;qwq(x) if(dist[o->to]>d+o->dist) dist[o->to]=d+o->dist,q.push(node(o->to,dist[o->to]));} } int main(){int N=read(),M=read(),s=read(),a=read(),b=read();rep(i,1,N){int u=read(),v=read(),d=read();add(u,v,d);}dijkstra(a);int ans=dist[s]+dist[b];dijkstra(b);ans=min(ans,dist[s]+dist[a]);printf("%d\n",ans);return 0; }

?bzoj2017：

神dp！！！用了和2101差不多的方法，還更神的壓掉一維。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=2005; int a[nmax],sum[nmax],f[nmax][nmax]; int main(){int n=read();dwn(i,n,1) a[i]=read();rep(i,1,n) sum[i]=sum[i-1]+a[i];rep(i,1,n) rep(j,1,n){f[i][j]=f[i][j-1];if(i>=j+j) f[i][j]=max(f[i][j],sum[i]-f[i-j-j][j+j]);if(i>=j+j-1)f[i][j]=max(f[i][j],sum[i]-f[i-j-j+1][j+j-1]);}printf("%d\n",f[n][1]);return 0; }

bzoj1673：

利用性質用dfs寫背包dp（加了一個剪枝；

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define ll long long #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=1005; ll a[nmax],sum[nmax],ans=0,N,C; void dfs(int cur,ll x){if(sum[cur-1]+x<=C) {ans=max(ans,sum[cur-1]+x);return ;}ans=max(ans,x);dwn(i,cur-1,1){x+=a[i];if(x<=C) dfs(i,x);x-=a[i];} } int main(){scanf("%lld%lld",&N,&C);rep(i,1,N) scanf("%lld",&a[i]),sum[i]=sum[i-1]+a[i];dfs(N+1,0);printf("%d\n",ans);return 0; }

bzoj2016：

裸的二分答案，輸出方案我原來那樣子寫在最后還更新了一下答案！最后還更新了答案！！！沒救了。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) #define ll long long ll read(){ll x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=50005; ll a[nmax],res[nmax],N,D; /*bool check(ll x){ll cur=0,tmp=0;rep(i,1,D){while(tmp<x&&cur<=N) tmp+=a[++cur],res[cur]=i;if(cur>N) return false;tmp/=2;}rep(i,cur+1,N) res[i]=D;return true; }*/ bool check(ll x){ll cur=0,tmp=0;rep(i,1,D){while(tmp<x&&cur<=N) tmp+=a[++cur];if(cur>N) return false;tmp/=2;}return true; } void get(ll x){ll cur=0,tmp=0;rep(i,1,D){while(tmp<x) tmp+=a[++cur],res[cur]=i;tmp/=2;}rep(i,cur+1,N) res[i]=D; } int main(){N=read(),D=read();ll l=0,r=0;rep(i,1,N) a[i]=read(),r+=a[i];ll mid,ans;while(l<=r){mid=(l+r)>>1;if(check(mid)) ans=mid,l=mid+1;else r=mid-1;}get(ans);printf("%lld\n",ans);rep(i,1,N) printf("%lld\n",res[i]);return 0; } /* 5 5 10 40 13 22 7 */

bzoj2015：

dijkstra。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=50005; const int maxn=200005; const int inf=0x7f7f7f7f; struct edge{int to,dist;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v,int d){pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++; }struct node{int x,dist;node(int x,int dist):x(x),dist(dist){};node(){};bool operator<(const node&rhs)const{return dist>rhs.dist;} }; priority_queue<node>q; int dist[nmax]; void dijkstra(){clr(dist,0x7f);dist[1]=0;q.push(node(1,0));int x,d;while(!q.empty()){node o=q.top();q.pop();x=o.x,d=o.dist;if(dist[x]!=d) continue;qwq(x) if(dist[o->to]>dist[x]+o->dist){dist[o->to]=dist[x]+o->dist;q.push(node(o->to,dist[o->to]));}} } int main(){int N=read(),M=read(),B=read(),u,v,d;rep(i,1,M){u=read(),v=read(),d=read();add(u,v,d);}dijkstra();rep(i,1,B){u=read(),v=read();printf("%d\n",dist[u]+dist[v]);}return 0; }

bzoj2014：

數據大唬人啊。。。然并卵。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) #define ll long long ll read(){ll _x=0;char _c=getchar();while(!isdigit(_c)) _c=getchar();while(isdigit(_c)) _x=_x*10+_c-'0',_c=getchar();return _x; } const int nmax=100005; struct node{ll w,c;bool operator<(const node&rhs)const{return c<rhs.c;} }; node nodes[nmax]; int main(){ll N=read(),m=read();rep(i,1,N) nodes[i].c=read(),nodes[i].w=read();sort(nodes+1,nodes+N+1);ll ans=0,tmp;rep(i,1,N){tmp=min(m/nodes[i].c,nodes[i].w);ans+=tmp;m-=nodes[i].c*tmp;}printf("%lld\n",ans);return 0; }

bzoj1689：

在枚舉水坑的時候枚舉木板就可以了

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) int _x;char _c; int read(){_x=0;_c=getchar();while(!isdigit(_c)) _c=getchar();while(isdigit(_c)) _x=_x*10+_c-'0',_c=getchar();return _x; } const int nmax=10005; struct node{int l,r;bool operator<(const node&rhs)const{return l<rhs.l||l==rhs.l&&r<rhs.r;} }; node nodes[nmax]; int main(){int N=read(),L=read();rep(i,1,N) nodes[i].l=read()+1,nodes[i].r=read();sort(nodes+1,nodes+N+1);int cnt=(nodes[1].r-nodes[1].l)/L+1,cur=nodes[1].l+cnt*L-1,tmp;rep(i,2,N){if(cur>=nodes[i].r) continue;if(nodes[i].l>cur){tmp=(nodes[i].r-nodes[i].l)/L+1;cur=nodes[i].l+tmp*L-1;cnt+=tmp;}else{tmp=(nodes[i].r-cur-1)/L+1;cur+=tmp*L;cnt+=tmp;}}printf("%d\n",cnt);return 0; }

bzoj1688：

N很小很明顯可以狀壓。然后位運算加括號加括號不然總是神奇的優先級啊。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) int _x;char _c; int read(){_x=0;_c=getchar();while(!isdigit(_c)) _c=getchar();while(isdigit(_c)) _x=_x*10+_c-'0',_c=getchar();return _x; } const int nmax=1005; int N,D,K,a[nmax],t[16]; bool pd(int x){int cnt=0;while(x) cnt+=x&1,x>>=1;return cnt<=K; } int main(){N=read(),D=read(),K=read();int u,v;rep(i,1,16) t[i]=1<<(i-1);rep(i,1,N){u=read();rep(j,1,u) v=read(),a[i]|=t[v];}int ans=0,sum,tmp=(1<<D)-1;rep(i,0,tmp) if(pd(i)){sum=0;rep(j,1,N) if((i|a[j])==i) sum++;ans=max(ans,sum);}printf("%d\n",ans);return 0; } /* 6 3 2 0 1 1 1 2 1 3 2 2 1 2 2 1 */

bzoj1684：

暴力！模擬！

#include<cstdio> double as(double x){return x<0?-x:x; } int main(){int N,D;scanf("%d%d",&N,&D);double ans=2.0,tmp=N*1.0/D;int ta,tb,temp;for(int i=1;i<=32767;i++){temp=tmp*i;if(as(tmp-temp*1.0/i)<ans&&i*N!=D*temp) ans=as(tmp-temp*1.0/i),ta=temp,tb=i;if(as(tmp-(temp+1)*1.0/i)<ans&&i*N!=D*(temp+1)) ans=as(tmp-(temp+1)*1.0/i),ta=temp+1,tb=i;}printf("%d %d\n",ta,tb);return 0; }

bzoj1682：

二分答案+spfa。。。。我。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) const int nmax=2005; const int maxn=20005; const int inf=0x7f7f7f7f; int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; }struct edge{int to,dist;bool flag;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v,int d){pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++; }int N,M; int q[nmax]; bool inq[nmax]; int dist[nmax]; bool spfa(){clr(inq,0);inq[1]=1;int x,l=1,r=1;q[1]=1;clr(dist,0x7f);dist[1]=0;while(l<=r){x=q[l];l++;qwq(x) if(o->flag&&dist[o->to]>dist[x]+o->dist){dist[o->to]=dist[x]+o->dist;if(!inq[o->to]) q[++r]=o->to,inq[o->to]=1;}}rep(i,1,N) if(dist[i]==inf) return false;return true; } bool check(int x){rep(i,1,N) qwq(i) if(o->dist<=x) o->flag=true;else o->flag=false;if(spfa()) return true;return false; }int main(){N=read(),M=read();int u,v,d,omax=0;rep(i,1,M){u=read(),v=read(),d=read();add(u,v,d);omax=max(omax,d);}int l=0,r=omax,mid,ans=0;while(l<=r){mid=(l+r)>>1;if(check(mid)) ans=mid,r=mid-1;else l=mid+1;}printf("%d\n",ans);return 0; }

bzoj1649：

找出每一段使用的次數就可以了。（分成兩半想。。不然會亂死。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define ll long long const int nmax=10005; int a[nmax]; int main(){int n;scanf("%d",&n);rep(i,1,n) scanf("%d",&a[i]);sort(a+1,a+n+1);ll ans=0;rep(i,1,n-1) ans+=(ll)i*(n-i)*(a[i+1]-a[i]);printf("%lld\n",ans<<1);return 0; }

bzoj1677：

背包dp。。。我。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) int a[25],dp[1000005]; int main(){int n;scanf("%d",&n);rep(i,0,25){if((1<<i)>n) break;a[++a[0]]=(1<<i);}dp[0]=1;rep(i,1,a[0]) rep(j,a[i],n)if((dp[j]+=dp[j-a[i]])>=1000000000)dp[j]-=1000000000;printf("%d\n",dp[n]);return 0; }

bzoj1671：

兩遍bfs就可以了（看清題意看清題意。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=1005; const int maxn=1000005; int map[nmax][nmax],oa[maxn],ob[maxn]; int dist[nmax][nmax],dis[nmax][nmax]; bool vis[nmax][nmax]; int q[maxn],Q[maxn]; int xx[5]={0,0,0,1,-1}; int yy[5]={0,1,-1,0,0}; int main(){int n,m,i,j,sa,sb,ta,tb,cnt=0;m=read(),n=read();rep(i,1,n) rep(j,1,m){map[i][j]=read();if(map[i][j]==2) sa=i,sb=j;else if(map[i][j]==3) ta=i,tb=j;else if(map[i][j]==4) oa[++cnt]=i,ob[cnt]=j;}int x,y,tx,ty,l,r;q[l=r=1]=sa;Q[l]=sb;vis[sa][sb]=1;map[ta][tb]=1;clr(dist,0x3f);dist[sa][sb]=0;while(l<=r){x=q[l];y=Q[l];l++;rep(i,1,4){tx=x+xx[i],ty=y+yy[i];if(tx<1||ty<1||tx>n||ty>m||map[tx][ty]==1||vis[tx][ty]) continue;vis[tx][ty]=1;dist[tx][ty]=dist[x][y]+1;q[++r]=tx;Q[r]=ty;}}q[l=r=1]=ta;Q[l]=tb;clr(vis,0);vis[ta][tb]=1;clr(dis,0x3f);dis[ta][tb]=0;while(l<=r){x=q[l];y=Q[l];l++;rep(i,1,4){tx=x+xx[i],ty=y+yy[i];if(tx<1||ty<1||tx>n||ty>m||map[tx][ty]==1||vis[tx][ty]) continue;vis[tx][ty]=1;dis[tx][ty]=dis[x][y]+1;q[++r]=tx;Q[r]=ty;}}int ans=0x7fffffff;rep(i,1,cnt) ans=min(ans,dist[oa[i]][ob[i]]+dis[oa[i]][ob[i]]);printf("%d\n",ans);return 0; }

bzoj1665：

轉換成最短路就可以了。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=10005; const int maxn=2000005; const int inf=0x7f7f7f7f; struct edge{int to,dist;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v,int d){pt->to=v;pt->dist=d;pt->next=head[u];head[u]=pt++;pt->to=u;pt->dist=d;pt->next=head[v];head[v]=pt++; } int H,N,s,t; queue<int>q; int dist[nmax],x[nmax],y[nmax]; bool inq[nmax]; void spfa(){clr(dist,0x7f);dist[0]=0;clr(inq,false);q.push(0);while(!q.empty()){int x=q.front();q.pop();inq[x]=false;qwq(x) if(dist[o->to]>dist[x]+o->dist){dist[o->to]=dist[x]+o->dist;if(!inq[o->to]) q.push(o->to),inq[o->to]=true;}}printf("%d\n",dist[t]-1); } int main(){H=read(),N=read(),s=0,t=N+1;rep(i,1,N) {x[i]=read(),y[i]=read();if(y[i]<=1000) add(s,i,1);if(H-y[i]<=1000) add(i,t,1);rep(j,1,i-1) if((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])<=1000000) add(i,j,1);}spfa();return 0; }

bzoj1672：

更區間覆蓋差不多。。改一下就可以了。挖坑！有比O(n2)的方法！

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=10005; const int inf=0x7f7f7f7f; struct node{int l,r,c;bool operator<(const node&rhs)const{return l<rhs.l||l==rhs.l&&r<rhs.r;} }; node nodes[nmax]; int dp[nmax]; int main(){int n=read(),s=read(),t=read();rep(i,1,n) {node &o=nodes[i];o.l=read(),o.r=read(),o.c=read();}sort(nodes+1,nodes+n+1);if(nodes[1].l>s) {printf("-1\n");return 0;}int ans=inf;clr(dp,0x7f);rep(i,1,n) {if(nodes[i].l<=s) dp[i]=nodes[i].c;rep(j,1,i-1) if(dp[j]!=inf&&nodes[i].l<=nodes[j].r+1)dp[i]=min(dp[i],dp[j]+nodes[i].c);if(nodes[i].r>=t) ans=min(ans,dp[i]);}if(ans==inf) ans=-1;printf("%d\n",ans);return 0; } /* 3 0 4 0 2 3 3 4 2 0 0 1 */

bzoj1664：

同前一題

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=10005; struct node{int l,r;bool operator<(const node&rhs)const{return l<rhs.l||l==rhs.l&&r<rhs.r;} }; node nodes[nmax]; int dp[nmax]; int main(){int n=read();rep(i,1,n) {node &o=nodes[i];o.l=o.r=read(),o.r+=read()-1;}sort(nodes+1,nodes+n+1);int ans=1;rep(i,1,n) {dp[i]=1;rep(j,1,i-1) if(nodes[i].l>nodes[j].r)dp[i]=max(dp[i],dp[j]+1);ans=max(ans,dp[i]);}printf("%d\n",ans);return 0; }

bzoj1662：

！！！打表第一個元素為0，打表第一個元素為0，打表第一個元素為0！！！哦一開始的表還打不完整，可以同時打兩個表然后一個輸出編號。那么就可以判斷打的完不完整啦。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define rep(i,s,t) for(int i=s;i<=t;i++) const int 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int t[2]; bool pd(int x){t[0]=t[1]=0;while(x) ++t[x&1],x>>=1;return t[0]>=t[1]; } int cal(int x){int tmp=x/200000,ans=0;rep(i,1,tmp) ans+=a[i];rep(i,x/200000*200000+1,x) if(pd(i)) ans++;return ans; } int main(){int n,m;scanf("%d%d",&n,&m);printf("%d\n",cal(m)-cal(n-1));return 0; }

bzoj1661：

O（n4）暴力枚舉

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=105; int map[nmax][nmax]; char s[nmax]; int main(){int n;scanf("%d",&n);rep(i,1,n){scanf("%s",s+1);rep(j,1,n) if(s[j]=='J') map[i][j]=2;else if(s[j]=='*') map[i][j]=1;else map[i][j]=0;}int x,y,ans=0;rep(i,1,n) rep(j,1,n) if(map[i][j]==2){rep(k,1,n) rep(t,1,n) if(map[k][t]){if(k==i&&t==j) continue;x=i-k,y=j-t;if(x*x+y*y<=ans) continue;if(i+y<=n&&k+y<=n&&j-x<=n&&t-x<=n&&i+y>0&&k+y>0&&j-x>0&&t-x>0&&map[i+y][j-x]==2&&map[k+y][t-x]==2)ans=x*x+y*y;x=k-i,y=t-j;if(i+y<=n&&k+y<=n&&j-x<=n&&t-x<=n&&i+y>0&&k+y>0&&j-x>0&&t-x>0&&map[i+y][j-x]==2&&map[k+y][t-x]==2)ans=x*x+y*y;}}printf("%d\n",ans);return 0; } /* 6J*J*********J***J******************* */

bzoj1660：

又是單調棧。。。我。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) #define ll long long const int nmax=80005; int a[nmax],s[nmax]; int main(){ll ans=0;int n;scanf("%d",&n);rep(i,1,n) scanf("%d",&a[i]);s[1]=1;ll r=1;rep(i,2,n){while(r&&a[s[r]]<=a[i]) r--;ans+=r;s[++r]=i;}printf("%lld\n",ans);return 0; }

bzoj1657：

還是單調棧。。。我。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int x;char c; int read(){x=0;c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=50005; int h[nmax],v[nmax],s[nmax],ans[nmax]; int main(){int N=read();rep(i,1,N) h[i]=read(),v[i]=read();s[1]=1;;int r=1;rep(i,2,N){while(r&&h[s[r]]<h[i]) ans[i]+=v[s[r]],r--;s[++r]=i;}s[1]=N;r=1;dwn(i,N-1,1){while(r&&h[s[r]]<h[i]) ans[i]+=v[s[r]],r--;s[++r]=i;}int res=0;rep(i,1,N) res=max(res,ans[i]);printf("%d\n",res);return 0; }

bzoj1655：

背包dp+高精度。。。看清楚數據范圍看清楚數據范圍看清楚數據范圍！！！

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) int dp[1005][1005]; void add(int x,int y){rep(i,1,dp[y][0]) dp[x][i]+=dp[y][i];dp[x][0]=max(dp[x][0],dp[y][0]);rep(i,1,dp[x][0]) if(dp[x][i]>=10) dp[x][i]-=10,dp[x][i+1]++;if(dp[x][dp[x][0]+1]) dp[x][0]++; } int main(){int n,k;scanf("%d%d",&n,&k);dp[0][0]=dp[0][1]=1;rep(i,1,k) rep(j,i,n) add(j,j-i);dwn(i,dp[n][0],1) printf("%d",dp[n][i]);return 0; }

bzoj1654：

tarjan即可

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<stack> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) #define qwq(x) for(edge *o=head[x];o;o=o->next) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=10005; const int maxn=50005; const int inf=0x7f7f7f7f; struct edge{int to;edge *next; }; edge edges[maxn],*pt=edges,*head[nmax]; void add(int u,int v){pt->to=v;pt->next=head[u];head[u]=pt++; } int sccno[nmax],pre[nmax],ans=0,dfs_clock=0,scc_cnt=0; stack<int>s; int dfs(int x){int lowu=pre[x]=++dfs_clock;s.push(x);qwq(x) {int to=o->to;if(!pre[to]) lowu=min(lowu,dfs(to));else if(!sccno[to]) lowu=min(lowu,pre[to]);}if(lowu==pre[x]){int sum=0;scc_cnt++;while(1){sum++;int X=s.top();s.pop();sccno[X]=scc_cnt;if(x==X) break;}if(sum!=1) ans++;}return lowu; } int main(){int N=read(),M=read(),u,v;rep(i,1,M) u=read(),v=read(),add(u,v);rep(i,1,N) if(!pre[i]) dfs(i);printf("%d\n",ans);return 0; }

bzoj1653：

N很小，求出所有全排列后利用楊輝三角形計算次數即可

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int a[11],dp[11][11]; int main(){int n,m,tmp;scanf("%d%d",&n,&m);rep(i,1,n) a[i]=i;dp[1][1]=1;rep(i,2,n) rep(j,1,i) dp[i][j]=dp[i-1][j]+dp[i-1][j-1];while(1){tmp=0;rep(i,1,n) tmp+=a[i]*dp[n][i];if(tmp==m) break;next_permutation(a+1,a+n+1);}printf("%d",a[1]);rep(i,2,n) printf(" %d",a[i]);printf("\n");return 0; }

bzoj1652：

和做過的差不多，可以壓掉一維不會空間沒炸就沒壓了。。。

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) const int nmax=2005; int a[nmax],f[nmax][nmax]; int main(){int n;scanf("%d",&n);rep(i,1,n) scanf("%d",&a[i]);rep(i,1,n) {for(int j=1;j+i-1<=n;j++){f[j][j+i-1]=max(f[j+1][j+i-1]+(n-i+1)*a[j],f[j][j+i-2]+(n-i+1)*a[j+i-1]);}}printf("%d\n",f[1][n]);return 0; }

bzoj1651：

差分序列即可

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=1000005; int a[nmax]; int main(){int N=read(),u,v,tot=-1,ans=0;rep(i,1,N) u=read(),v=read(),a[u]+=1,a[v+1]-=1,tot=max(tot,v);rep(i,1,tot) ans=max(ans,a[i]+=a[i-1]);printf("%d\n",ans);return 0; }

?bzoj1649：

跟背包dp差不多

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)); int read(){int x=0;char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) x=x*10+c-'0',c=getchar();return x; } const int nmax=1005; int dp[nmax][nmax]; struct node{int s,t,w,c;bool operator<(const node&rhs)const{return s<rhs.s||s==rhs.s&&t<rhs.t;} }; node nodes[nmax*10]; int main(){int L=read(),N=read(),B=read();rep(i,1,N) {node &o=nodes[i];o.s=o.t=read(),o.t+=read(),o.w=read(),o.c=read();}sort(nodes+1,nodes+N+1);clr(dp,-1);dp[0][0]=0;rep(i,1,N) dwn(j,B,nodes[i].c) if(dp[nodes[i].s][j-nodes[i].c]!=-1)dp[nodes[i].t][j]=max(dp[nodes[i].s][j-nodes[i].c]+nodes[i].w,dp[nodes[i].t][j]);int ans=-1;rep(i,1,B) if(dp[L][i]!=-1) ans=max(ans,dp[L][i]);printf("%d\n",ans);return 0; }

?

轉載于:https://www.cnblogs.com/fighting-to-the-end/p/5760043.html

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