The Bottom of a Graph 强接通分量加缩点

The Bottom of a Graph 强连通分量加缩点

/*题意比较晦涩，大致就是求一个图缩点后出度为0的点的个数。*/

#include <stdio.h>
#include <cstring>
#include <vector>
#include <iostream>
using namespace std;
vector<int> e[5010];
int dfn[5001];
int low[5001];
int stack[5001];
bool instack[5001];
int belong[5001];
int out[5001];
int n,m,num,top,cnt;
void tarjan(int u)
{
    int j;
    stack[top++]=u;
    low[u]=dfn[u]=++num;
    instack[u]=true;
    for(int i=0; i<e[u].size(); i++)
    {
        int v=e[u][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[v],low[u]);
        }
        else if(instack[v])
            low[u]=min(dfn[v],low[u]);
    }
    if(low[u]==dfn[u])
    {
        cnt++;
        do
        {
            j=stack[--top];
            instack[j]=false;
            belong[j]=cnt;
        }
        while(j!=u);
    }
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        if(n==0) break;
        scanf("%d",&m);
        int a,b;
        for(int i=1; i<=n; i++)
            e[i].clear();
        top=0;
        num=0;
        cnt=0;
        while(m--)
        {
            scanf("%d%d",&a,&b);
            e[a].push_back(b);
        }
        memset(low,0,sizeof(low));
        memset(dfn,0,sizeof(dfn));
        memset(instack,false,sizeof(instack));
        memset(belong,0,sizeof(belong));
        for(int i=1; i<=n; i++)
        {
            out[i]=0;
            if(!dfn[i])
                tarjan(i);
        }
        for(int i=1; i<=n; i++)
        {
            int len=e[i].size();
            for(int j=0; j<len; j++)
            {
                if(belong[i]!=belong[e[i][j]])
                    out[belong[i]]++;
            }
        }
        int k=0;
        for(int i=1; i<=n; i++)
        {
            if(out[belong[i]]==0)
            {
                if(k++) printf(" ");
                printf("%d",i);
            }
        }
        printf("\n");
    }
    return 0;
}