好久没写题解了＝ ＝这次是bzoj 1051

唉＝ ＝这道题我都想到了tarjan缩点，但是没有想到最后一步啊＝ ＝我们很容易想到反向建边然后缩点，这时候我们看由多少个联通块的入度为0，如果为1个，那就输出这个块的大小，否则输出0；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 10010;
const int maxe = 50010;

struct edge {
    int t;
    edge* next;
}e[maxe * 2], *head[maxn]; int ne = 0;

int n, m;

void addedge(int f, int t) {
    e[ne].t = t, e[ne].next = head[f]; head[f] = e + ne ++;
}

void read() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; ++ i) {
        int u, v;
        scanf("%d%d", &u, &v);
        addedge(v, u);
    }
}

int dfs[maxn], back[maxn];
int s[maxn], top = 0, Index = 0;
bool vis[maxn]; int num = 0;
int size[maxn];
int belong[maxn];

void tarjan(int now) {
    dfs[now] = back[now] = ++ Index; vis[now] = true;
    s[++ top] = now;
    for(edge* p = head[now]; p; p = p-> next) {
        if(!dfs[p-> t]) tarjan(p-> t), back[now] = min(back[now], back[p-> t]);
        else if(vis[p-> t] && back[now] > dfs[p-> t]) { 
            back[now] = dfs[p-> t];
        }
    }
    if(dfs[now] == back[now]) {
        ++ num;
        while(1) {
            belong[s[top]] = num; 
            size[num] ++;
            vis[s[top]] = false;
            if(s[top] == now) break;
            top --;
        }
        top --;
    }
}

int f[maxn]; int in[maxn];
bool have[maxn];

void sov() {
    memset(vis, false, sizeof(false));
    memset(size, 0, sizeof(size));
    memset(have, 0, sizeof(have));
    for(int i = 1; i <= n; ++ i) {
        if(!dfs[i]) tarjan(i);
    }
    memset(in, 0, sizeof(in));
    int tol = 0;
    for(int i = 1; i <= n; ++ i) {
        for(edge* p = head[i]; p; p = p-> next) {
            if(belong[p-> t] != belong[i]) {
                in[belong[p-> t]] ++;
            }
        }
    }
    int pos = 0;
    for(int i = 1; i <= num; ++ i) {
        if(in[i] == 0) tol ++, pos = i;
    }
    if(tol == 1) printf("%d\n", size[pos]);
    else printf("0\n");
}

int main() {
    //freopen("test.in", "r", stdin);
    read(); sov();
    return 0;
}