POJ 2553The Bottom of a Graph(Tarjan)

题意：我的英语可是一流的。。。（水）。。。没看懂题，看图看明白的。输出出度为0的团的所有元素。（猜的。。过了，说明就猜对了）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 50009;
const int M = N*N;
struct LT{
    int nex,to;
}L[N];
int F[N],cnt;
void add(int f,int t)
{
    L[cnt].nex = F[f];
    L[cnt].to = t;
    F[f] = cnt++;
}
int n,m;
int dfn[N],low[N],col[N],color,ind;
bool post[N];
stack<int> S;
void tarjan(int k)
{
    dfn[k] = low[k] = ind++;
    post[k] = true;
    S.push(k);
    for(int i=F[k];i;i=L[i].nex)
    {
        int to = L[i].to;
        if(!dfn[to])
        {
            tarjan(to);
            low[k] = min(low[k],low[to]);
        }else if(post[to]&&low[k]>dfn[to])
        {
            low[k] = dfn[to];
        }
    }
    if(low[k]==dfn[k])
    {
        color++;int i;
        for(i=S.top(),S.pop();i!=k;i=S.top(),S.pop())
        {
            col[i] = color,post[i] = false;
        }
        col[i]=color,post[i] = false;
    }
}
bool ou[N];
void solve()
{
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(ou,false,sizeof(ou));
    color = 1;ind = 1;
    for(int i =1;i<=n;i++)
    if(!dfn[i]) tarjan(i);
    for(int i=1;i<=n;i++)
    {
        for(int j=F[i];j;j=L[j].nex)
        {
            if(col[i]!=col[L[j].to])
            {
                ou[col[i]]=true;
            }
        }
    }
    bool o = false;
    for(int i=1;i<=n;i++)
    if(!ou[col[i]])
    {
        if(o) printf(" ");o = true;
        printf("%d",i);
    }
    printf("\n");
}
int main()
{
    freopen("in.txt","r",stdin);
    while(~scanf("%d",&n)&&n)
    {
        scanf("%d",&m);
        memset(F,0,sizeof(F));cnt =1;
        int f,t;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&f,&t);
            add(f,t);
        }
        solve();
    }
    return 0;
}