POJ2553 The Bottom of a Graph（强连通分量+缩点）

题目是问，一个有向图有多少个点v满足∀w∈V:(v→w)⇒(w→v)。

把图的强连通分量缩点，那么答案显然就是所有出度为0的点。

用Tarjan找强连通分量：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 5555
#define MAXM 5555*5555
struct Edge{
    int u,v,next;
}edge[MAXM];
int NE,head[MAXN];
void addEdge(int u,int v){
    edge[NE].u=u; edge[NE].v=v; edge[NE].next=head[u];
    head[u]=NE++;
}

int bn,belong[MAXN],stack[MAXN],top;
bool instack[MAXN];
int dn,dfn[MAXN],low[MAXN];
void dfs(int u){
    dfn[u]=low[u]=++dn;
    stack[++top]=u; instack[u]=1;
    for(int i=head[u]; i!=-1; i=edge[i].next){
        int v=edge[i].v;
        if(dfn[v]==0){
            dfs(v);
            low[u]=min(low[u],low[v]);
        }else if(instack[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u]){
        int v; ++bn;
        do{
            v=stack[top--];
            instack[v]=0;
            belong[v]=bn;
        }while(u!=v);
    }
}

int deg[MAXN];
int main(){
    int n,m,a,b;
    while(~scanf("%d",&n) && n){
        NE=0;
        memset(head,-1,sizeof(head));
        scanf("%d",&m);
        while(m--){
            scanf("%d%d",&a,&b);
            addEdge(a,b);
        }
        top=dn=bn=0;
        memset(dfn,0,sizeof(dfn));
        memset(instack,0,sizeof(instack));
        for(int i=1; i<=n; ++i){
            if(dfn[i]==0) dfs(i);
        }
        memset(deg,0,sizeof(deg));
        for(int i=0; i<NE; ++i){
            int u=belong[edge[i].u],v=belong[edge[i].v];
            if(u==v) continue;
            ++deg[u];
        }
        bool first=1;
        for(int i=1; i<=n; ++i){
            if(deg[belong[i]]==0){
                if(first) first=0;
                else putchar(' ');
                printf("%d",i);
            }
        }
        putchar('
');
    }
    return 0;
}