bzoj 1051: [HAOI2006]受欢迎的牛

tarjan强连通分量求缩点重构图，出度为0的点若只有一个则输出其代表强连通分量的大小，否则无解。

因为一旦出度为0就没人被他认为长得帅

/*
ID:WULALA
PROB:bzoj1051 
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <fstream>
#include <ctime>
#define N 10008
#define M 50008
#define mod
#define mid(l,r) ((l+r) >> 1)
#define INF 0x7ffffff
using namespace std;

int n,m,dfn[N],low[N],que[N],head[N],cnt,h[N],scc,belong[N],ans,hav[N],r;
bool vis[N],inq[N],flag;

struct WULALA
{
    int node,next;
}e[M],d[M];

void init()
{
    scanf("%d%d",&n,&m);
    for (int i = 1;i <= m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        e[i].node = y;
        e[i].next = head[x];
        head[x] = i;
    }
}

void dfs(int a)
{
    vis[a] = true;
    low[a] = dfn[a] = ++cnt;
    inq[a] = true; que[++r] = a;
    int c = head[a];
    while(c)
    {
        if (!vis[e[c].node]) 
        {
            dfs(e[c].node);
            low[a] = min(low[a],low[e[c].node]);
        }
        else if (inq[e[c].node]) low[a] = min(low[a],dfn[e[c].node]);//要判是否在队列里！！ 
        c = e[c].next;
    }
    if (low[a] == dfn[a])
    {
        belong[a] = ++scc;
        hav[scc] = 1;
        while (que[r] != a)
        {
            belong[que[r]] = scc;
            inq[que[r]] = false;
            ++hav[scc];
            --r;
        }
        inq[que[r]] = false;//!!!
        --r;//!!!
    }
}

void rebuild()
{
    cnt = 0;
    for (int i = 1;i <= n;i++)
    {
        int c = head[i];
        while(c)
        {
            if (belong[i] != belong[e[c].node])
            {
                d[++cnt].node = belong[e[c].node];    
                d[cnt].next = h[belong[i]];
                h[belong[i]] = cnt;
            }
            c = e[c].next;
        }
    }
}

void tarjan()
{

    for (int i = 1;i <= n;i++)
        if (!vis[i]) dfs(i);
    rebuild();
}

void work()
{
    for (int i = 1;i <= scc;i++)
        if (!h[i])
        {
            if (ans)
            {
                ans = 0;
                return;
            }
            else ans = hav[i];
        }
}

int main()
{
    init();
    tarjan();
    work();
    printf("%d
",ans);
    return 0;
}