hdu 1269 强连通图的判定

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给你一个图，判断是不是强连通图。
强连通图的强连通分量就是1.套了下kuangbin的模板。
单向边，注意init

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int MAXN = 10010;//点数const int MAXM = 100010;//边数struct Edge{int to,next;}edge[MAXM];int head[MAXN],tot;int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~sccint Index,top;int scc;//强连通分量的个数bool Instack[MAXN];int num[MAXN];//各个强连通分量包含点的个数，数组编号1~scc//num数组不一定需要，结合实际情况void addedge(int u,int v){edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;}void Tarjan(int u){int v;Low[u] = DFN[u] = ++Index; Stack[top++] = u; Instack[u] = true;for(int i = head[u];i != -1;i = edge[i].next){v = edge[i].to;if( !DFN[v] ){Tarjan(v);if( Low[u] > Low[v] )Low[u] = Low[v];}else if(Instack[v] && Low[u] > DFN[v]) Low[u] = DFN[v];}if(Low[u] == DFN[u]){scc++;do{v = Stack[--top]; Instack[v] = false; Belong[v] = scc; num[scc]++;}while( v != u);}}void solve(int N){memset(DFN,0,sizeof(DFN)); memset(Instack,false,sizeof(Instack)); memset(num,0,sizeof(num));Index = scc = top = 0;for(int i = 1;i <= N;i++)if(!DFN[i])Tarjan(i);}void init(){tot = 0;memset(head,-1,sizeof(head));}int n,m,u,v;int main(){    while(~scanf("%d%d",&n,&m)){        if(n==0&&m==0)break;        init();        for(int i=0;i<m;i++){            scanf("%d%d",&u,&v);            addedge(u,v);        }        solve(n);        if(scc==1){            printf("Yes\n");        }        else printf("No\n");    }    return 0;}

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