Jzoj3899 逻辑的连通性

题意：给你一个图，求出每个节点所在的联通快的大小总和

显然就是tarjan找强连通分量了，没啥好说

//#pragma GCC optimize("O3")//#pragma G++ optimize("O3")#include<stdio.h>#include<string.h>#include<algorithm>#include<stack>using namespace std;struct Edge{ int v,nt; } G[600010];int h[50010]={0},C=0;int dfn[50010],low[50010],clk=0;int n,m,col[50010],cnt=0,b[50010];bool in[50010]; stack<int> st;inline void adj(int x,int y){G[++C]=(Edge){y,h[x]}; h[x]=C;}void dfs(int u){dfn[u]=low[u]=++clk;st.push(u); in[u]=1;for(int v,i=h[u];i;i=G[i].nt)if(!dfn[v=G[i].v]){dfs(v);low[u]=min(low[u],low[v]);} else if(in[v]) low[u]=min(low[u],dfn[v]);if(low[u]==dfn[u]){++cnt; for(int v;(v=st.top())!=u;){col[v]=cnt; b[cnt]++;in[v]=0; st.pop();}col[u]=cnt; b[cnt]++; in[u]=0; st.pop();}}int main(){long long ans=0;scanf("%d%d",&n,&m);for(int x,y,i=0;i<m;++i){scanf("%d%d",&x,&y);adj(x,y);}for(int i=1;i<=n;++i) if(!dfn[i]) dfs(i);for(int i=1;i<=cnt;++i) ans+=(long long)b[i]*(b[i]-1)>>1;printf("%lld\n",ans);}