无向图缩点:tarjan点双与边双缩点(模板)
生活随笔
收集整理的這篇文章主要介紹了
无向图缩点:tarjan点双与边双缩点(模板)
小編覺得挺不錯的,現在分享給大家,幫大家做個參考.
e-DCC邊雙縮點:(用之前記得init) cnt1是從2開始的
邊雙縮點可以將無向圖轉換為一個森林,如果原圖保證連通的話,那么可以轉換為一棵樹
const int N=1e4+100;const int M=1e5+100;struct Egde {int to,next; }edge1[M],edge2[M];int head1[N],head2[N],low[N],dfn[N],c[N],num,cnt1,cnt2,dcc,n,m;bool bridge[M];void addedge1(int u,int v) {edge1[cnt1].to=v;edge1[cnt1].next=head1[u];head1[u]=cnt1++; }void addedge2(int u,int v) {edge2[cnt2].to=v;edge2[cnt2].next=head2[u];head2[u]=cnt2++; }void tarjan(int u,int in_edge) {dfn[u]=low[u]=++num;for(int i=head1[u];i!=-1;i=edge1[i].next){int v=edge1[i].to;if(!dfn[v]){tarjan(v,i);low[u]=min(low[u],low[v]);if(low[v]>dfn[u])bridge[i]=bridge[i^1]=true;}else if(i!=(in_edge^1))low[u]=min(low[u],dfn[v]);} }void dfs(int u) {c[u]=dcc;for(int i=head1[u];i!=-1;i=edge1[i].next){int v=edge1[i].to;if(c[v]||bridge[i])continue;dfs(v);} }void solve() {for(int i=1;i<=n;i++)//找橋 if(!dfn[i])tarjan(i,0);for(int i=1;i<=n;i++)//縮點 if(!c[i]){dcc++;dfs(i);} }void build()//縮點+連邊 {solve();for(int i=2;i<cnt1;i+=2){int u=edge1[i^1].to;int v=edge1[i].to;if(c[u]==c[v])continue;addedge2(c[u],c[v]);addedge2(c[v],c[u]);} }void init() {cnt1=2;cnt2=num=dcc=0;memset(head2,-1,sizeof(head2));memset(head1,-1,sizeof(head1));memset(low,0,sizeof(low));memset(dfn,0,sizeof(dfn));memset(bridge,false,sizeof(bridge));memset(c,0,sizeof(c)); }v-DCC點雙縮點:(用之前記得init)
const int N=1e4+100;const int M=1e5+100;struct Egde {int to,next; }edge1[M],edge2[M];int head1[N],head2[N],low[N],dfn[N],c[N],Stack[N],new_id[N],num,cnt,cnt1,cnt2,tot,root,top,n,m;bool cut[N];vector<int>dcc[N];void addedge1(int u,int v) {edge1[cnt1].to=v;edge1[cnt1].next=head1[u];head1[u]=cnt1++; }void addedge2(int u,int v) {edge2[cnt2].to=v;edge2[cnt2].next=head2[u];head2[u]=cnt2++; }void tarjan(int u) {dfn[u]=low[u]=++num;Stack[++top]=u;if(u==root&&head1[u]==-1){dcc[++cnt].push_back(u);return;}int flag=0;for(int i=head1[u];i!=-1;i=edge1[i].next){int v=edge1[i].to;if(!dfn[v]){tarjan(v);low[u]=min(low[u],low[v]);if(low[v]>=dfn[u]){flag++;if(u!=root||flag>1)cut[u]=true;cnt++;int x;do{x=Stack[top--];dcc[cnt].push_back(x);}while(x!=v);dcc[cnt].push_back(u);}}elselow[u]=min(low[u],dfn[v]);} }void solve() {for(int i=1;i<=n;i++)//找割點+縮點 if(!dfn[i]){root=i;tarjan(i);} }void build()//縮點+連邊 {solve();num=cnt;for(int i=1;i<=n;i++)if(cut[i])new_id[i]=++num;for(int i=1;i<=cnt;i++)for(int j=0;j<dcc[i].size();j++){int x=dcc[i][j];if(cut[x]){addedge2(i,new_id[x]);addedge2(new_id[x],i);}elsec[x]=i;} }void init() {for(int i=0;i<N;i++)dcc[i].clear();cnt=cnt2=cnt1=num=tot=top=0;memset(head2,-1,sizeof(head2));memset(head1,-1,sizeof(head1));memset(low,0,sizeof(low));memset(dfn,0,sizeof(dfn));memset(cut,false,sizeof(cut));memset(c,0,sizeof(c));memset(Stack,0,sizeof(Stack));memset(new_id,0,sizeof(new_id)); }?
總結
以上是生活随笔為你收集整理的无向图缩点:tarjan点双与边双缩点(模板)的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: POJ - 2186 Popular C
- 下一篇: POJ - 3189 Steady Co