P2341 [USACO03FALL][HAOI2006]受欢迎的牛 G

题目传送门！！

经典强联通题目。

对于一个强联通，如果其出度为0，则其中所有的奶牛都为明星。

注意，如果出现两个以上出度为0的强联通，则无法做到“被全部喜欢”， 应该特判输出“0”。

#include <bits/stdc++.h>
using namespace std;
#define N 100010
#define isdigit(c) ((c)>='0'&&(c)<='9')

int read(){
    int x = 0, s = 1;
    char c = getchar();
    while(!isdigit(c)){
        if(c == '-')s = -1;
        c = getchar();
    }
    while(isdigit(c)){
        x = (x << 1) + (x << 3) + (c ^ '0');
        c = getchar(); 
    } 
    return x * s;
} 

int dfn[N], low[N], scc[N];
int stac[N], top = 0;
bool vis[N]; 

struct node{
    int u, v, next;
} t[N];
int f[N];
int out[N];

int bian = 0;
inline void add(int u, int v){
    t[++bian] = (node){u, v, f[u]};
    f[u] = bian;
    return ;
}

int cac = 0, cnt = 0;
void tarjan(int now){  /*tarjan 板子*/
    dfn[now] = low[now] = ++cac;
    stac[++top] = now;
    vis[now] = 1;
    for(int i = f[now]; i; i = t[i].next){
        int v = t[i].v;
        if(!dfn[v]){
            tarjan(v);
            low[now] = min(low[now], low[v]);
        }
        else if(vis[v]) low[now] = min(low[now], dfn[v]);
    }
    if(low[now] == dfn[now]){
        int cur;
        cnt++;
        do{
            cur = stac[top--];
            vis[cur] = 0;
            scc[cur] = cnt;
        } while(cur != now);
    }
    return ;
}

int main(){
    int n = read(), m = read();
    for(int i = 1;i <= m; i++){
        int x = read(), y = read();
        add(x, y);
    }
    for(int i = 1;i <= n; i++)
        if(!dfn[i])  tarjan(i);
    for(int i = 1;i <= bian; i++){
        int u = t[i].u, v = t[i].v;
        if(scc[u] != scc[v])
            out[scc[u]] ++;   /*记录出度*/
    }
    int flag = 0, id = 0;
    for(int i = 1;i <= cnt; i++){
        if(out[i] == 0){
            flag++;
            id = i;
        }
    }
    int ans = 0;
    for(int i = 1;i <= n; i++)
        ans += scc[i] == id;
    if(flag > 1) printf("0");
    else printf("%d
", ans);
    return  0;
}