B.?字符路徑

給一個(gè)含n個(gè)點(diǎn)m條邊的有向無(wú)環(huán)圖（允許重邊，點(diǎn)用1到n的整數(shù)表示），每條邊上有一個(gè)字符，問(wèn)圖上有幾條路徑滿足路徑上經(jīng)過(guò)的邊上的字符組成的的字符串去掉空格后以大寫字母開頭，句號(hào) '.' 結(jié)尾，中間都是小寫字母，小寫字母可以為0個(gè)。

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dp[x][0]為全空格的方案, dp[x][1]為空格加字母的方案, dp[x][2]為合法路徑數(shù).

#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;} //headconst int N = 1e6+10; int n, m, deg[N], dp[N][3]; struct _ {int to;char w;}; vector<_> g[N]; queue<int> q;int main() {scanf("%d%d", &n, &m);REP(i,1,m) {int u, v;char w;scanf("%d%d %c", &u, &v, &w);g[u].pb({v,w});++deg[v];}REP(i,1,n) if (!deg[i]) q.push(i);unsigned ans = 0;while (q.size()) {int u = q.front(); q.pop();ans += dp[u][2];for (_ e:g[u]) {int v = e.to;if (isupper(e.w)) {dp[v][1] += dp[u][0]+1;}else if (islower(e.w)) { dp[v][1] += dp[u][1];}else if (e.w=='_') {dp[v][0] += dp[u][0]+1;dp[v][1] += dp[u][1];dp[v][2] += dp[u][2];}else if (e.w=='.') {dp[v][2] += dp[u][1];}if (!--deg[v]) q.push(v);}}printf("%u\n", ans); }

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D.整數(shù)序列

大意: 區(qū)間加, 求區(qū)間sin和.

線段樹維護(hù), 注意爆int

#include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;}) using namespace std; typedef long long ll; const int N = 2e5+10; int n, m, a[N]; struct _ {double c, s;ll tag;void upd(int x) {tie(c,s) = pair<double,double>(c*cos(x)-s*sin(x),c*sin(x)+s*cos(x)); tag += x;}_ operator + (const _ & rhs) const {return {c+rhs.c,s+rhs.s,0ll};} } tr[N<<2];void pd(int o) {if (tr[o].tag) {tr[lc].upd(tr[o].tag),tr[rc].upd(tr[o].tag);tr[o].tag = 0;} } void build(int o, int l, int r) {if (l==r) tr[o]={cos(a[l]),sin(a[l]),0ll};else build(ls),build(rs),tr[o]=tr[lc]+tr[rc]; } void update(int o, int l, int r, int ql, int qr, int v) { if (ql<=l&&r<=qr) return tr[o].upd(v);pd(o);if (mid>=ql) update(ls,ql,qr,v);if (mid<qr) update(rs,ql,qr,v);tr[o]=tr[lc]+tr[rc]; } double query(int o, int l, int r, int ql, int qr) {if (ql<=l&&r<=qr) return tr[o].s;pd(o);double ans = 0;if (mid>=ql) ans += query(ls,ql,qr);if (mid<qr) ans += query(rs,ql,qr);return ans; }int main() {scanf("%d", &n);REP(i,1,n) scanf("%d", a+i);build(1,1,n);scanf("%d", &m);while (m--) {int op,l,r,v;scanf("%d%d%d",&op,&l,&r);if (op==1) {scanf("%d", &v);update(1,1,n,l,r,v);}else if (op==2) printf("%.1lf\n", query(1,1,n,l,r));} }

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E. 骨牌覆蓋

給定棋盤, 一些格子有擋板, 求骨牌擺放方案數(shù).

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轉(zhuǎn)載于:https://www.cnblogs.com/uid001/p/10971022.html

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