tarjan算法求连通分量+缩点后重新建图

#include<bits/stdc++.h>using namespace std;const int maxn=2e4+7;vector<int> g[maxn];vector<int> newp[maxn];int dfn[maxn],low[maxn],sccno[maxn];int idx=0;bool visit[maxn];stack<int> s;void tarjan(int u){    dfn[u]=low[u]=++idx;    s.push(u);    visit[u]=true;    int v;    for(int i=0;i<g[u].size();++i){        v=g[u][i];        if(dfn[v]==0){            tarjan(v);            low[u]=min(low[u],low[v]);        }else if(visit[v])            low[u]=min(low[u],dfn[v]);    }    if(dfn[u]==low[u])        do{            v=s.top();            s.pop();            sccno[v]=u;//缩点            visit[v]=false;        }while(u!=v);}void suodian(int n){    int v,u;    for(int i=1;i<=n;++i){        for(int j=0;j<g[i].size();++j){            u=g[i][j];            if(sccno[i]==u)                continue;            if(w[u]==0)                continue;            if(u!=sccno[u]){                w[sccno[u]]+=w[u];                w[u]=0;            }else                newp[sccno[i]].push_back(u);        }    }}int main(){    ios::sync_with_stdio(false);    //freopen("in.txt","r",stdin);    int n,m,u,v;    memset(dfn,0,sizeof(dfn));    cin>>n>>m;    for(int i=1;i<=n;++i)        cin>>w[i];    for(int i=1;i<=m;++i){        cin>>u>>v;        g[u].push_back(v);    }    for(int i=1;i<=n;++i)        sccno[i]=i;    tarjan(1);    suodian(n);    return 0;}