【luogu P2341 [HAOI2006]受欢迎的牛】 题解

题解报告：https://www.luogu.org/problemnew/show/P2341
我们把图中的强连通分量缩点，然后只有出度为0的牛是受欢迎的，这样如果出度为0的牛只有一个，说明受所有牛欢迎。否则出度为0只是受一些牛欢迎。

#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 10;
struct edge{
    int next, to, from, len;
}e[maxn<<2];
int head[maxn], cnt;
int n, m, color[maxn], tong[maxn], du[maxn], num, tim, ans, tmp;
int dfn[maxn], low[maxn];
bool vis[maxn];
stack<int> s;
void add(int u, int v)
{
    e[++cnt].from = u;
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt;
}
void tarjan(int x)
{
    dfn[x] = low[x] = ++tim;
    s.push(x); vis[x] = 1;
    for(int i = head[x]; i != -1; i = e[i].next)
    {
        int v = e[i].to;
        if(!dfn[v])
        {
            tarjan(v);
            low[x] = min(low[x], low[v]);
        }
        else if(vis[v])
        {
            low[x] = min(low[x], low[v]);
        }
    }
    if(dfn[x] == low[x])
    {
        color[x] = ++num;
        vis[x] = 0;
        while(s.top() != x)
        {
            color[s.top()] = num;
            vis[s.top()] = 0;
            s.pop(); 
        }
        s.pop();
    }
}
int main()
{
    memset(head, -1, sizeof(head));
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= m; i++)
    {
        int u, v;
        scanf("%d%d",&u,&v);
        add(u,v);
    }
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= n; i++)
    {
        for(int j = head[i]; j != -1; j = e[j].next)
        {
            int v = e[j].to;
            if(color[v] != color[i])
            {
                du[color[i]]++;
            }
        }
        tong[color[i]]++;
    }
    for(int i = 1; i <= num; i++)
    {
        if(du[i] == 0)
        {
            tmp++;
            ans = tong[i];
        }
    }
    if(tmp == 0) 
    {
        printf("0");
        return 0;
    }
    if(tmp == 1)
    {
        printf("%d",ans);
        return 0;
    }
    if(tmp > 1)
    {
        printf("0");
        return 0;
    }
}