【tarjan缩点】受欢迎的牛

P2341 [HAOI2006]受欢迎的牛 

缩点令我快乐哈哈哈哈哈哈哈哈哈哈哈哈哈哈哈

一遍过编译一遍A哈哈哈哈哈哈开心快乐哈哈哈哈哈哈哈哈哈哈

好吧其实很简单

因为牛的爱慕具有传递性（爱牛及牛？？

所以出度为0的强连通分量内牛的个数即为答案， 但如果存在两个及两个以上的出度为0的强连通分量， ans = 0， 因为这些强连通之间不连通qaq

差不多就是这样啦

noip2018 加油ヾ(◍°∇°◍)ﾉﾞ

#include<cstdio>
#include<iostream>
#define ri register int
#define ll long long
#define maxn 10010
#define maxm 50050
using namespace std;
int n, m, num = 0, top = 0, tim = 0, cnt = 0, num2 = 0;
int node[maxn], head[maxm], head2[maxm],sd[maxn], chu[maxn], sta[maxn], low[maxn], dfn[maxn], x[maxm], y[maxm];
bool vis[maxn];
struct Edge {
    int nxt, to;
}e[maxn << 1], edge[maxm << 1];
int read() {
    char ch = getchar(); int x = 0, f = 1;
    while(ch > '9' || ch < '0') {if(ch == '-') f = -1; ch = getchar();}
    while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0' ; ch = getchar();}
    return x * f;
}
void add(int from, int to) {
    e[++num].nxt = head[from];
    e[num].to = to;
    head[from] = num;
}
void readd(int from, int to) {
    chu[from]++;
    edge[++num2].nxt = head2[from];
    edge[num2].to = to;
    head2[from] = num2;
}
void tarjan(int x) {
    low[x] = dfn[x] = ++tim;
    sta[++top]  = x;
    vis[x] = 1;
    for(int i = head[x]; i; i = e[i].nxt) {
        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], dfn[v]);
    }
    if(dfn[x] == low[x]) {
        cnt++;
        while(x != sta[top+1]) {
            sd[sta[top]] = cnt;
            vis[sta[top]] = 0;
            node[cnt]++;
            top--;
        }
    }
}
void search() {
    int flag = 0, ans = 0;
    for(int i = 1; i <= cnt; i++) {
        if(!chu[i]) {
            if(!flag) {
                ans = node[i];
                flag = 1;
            }
            else {
                flag = 2;
                break;
            }
        }
    }
    if(flag == 1) {
        printf("%d", ans);
        return;
    }
    else {
        printf("0");
        return;
    }
}
int main() {
    scanf("%d%d", &n, &m);
    for(ri i = 1; i <= m; i++) {
        x[i] = read(), y[i] = read();
        add(x[i], y[i]);
    }
    for(ri i = 1; i <= n; i++) 
        if(!dfn[i]) tarjan(i);
    for(ri i = 1; i <= m; i++) {
        int u = sd[x[i]], v = sd[y[i]];
        if(u != v) readd(u, v);
    }
    search();
    return 0;
}