【BZOJ1051】[HAOI2006]受欢迎的牛

【BZOJ1051】[HAOI2006]受欢迎的牛

题面

bzoj

洛谷

题解

假如\(A\)喜欢\(B\)就连一条\(A\)到\(B\)的边

然后缩点，如果图不连通就\(Impossible\)

否则输出出度为\(0\)的环的大小

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
inline int gi(){
    register int data=0, w = 1;
    register char ch=0;
    while (ch != '-' && (ch > '9' || ch < '0')) ch = getchar();
    if (ch == '-') w = -1 , ch = getchar();
    while (ch >= '0' && ch <= '9') data = (data << 1) + (data << 3) + (ch ^ 48), ch = getchar();
    return w * data;
}
#define MAX_N 10005
#define MAX_M 50005
struct Edge {
    int to, next;
} e[MAX_M];
int h[MAX_N], cnt;
int n, m;
void Add_Edge(int u, int v) {
    e[cnt].to = v, e[cnt].next = h[u], h[u] = cnt++;
}
int tim = 0, num = 0, top = 0;
int low[MAX_N], dfn[MAX_N], color[MAX_N], st[MAX_N];
int tot[MAX_N];
void tarjan(int x) {
    low[x] = dfn[x] = ++tim;
    st[++top] = x;
    for (register int i = h[x]; ~i; i = e[i].next) {
        if (!dfn[e[i].to]) {
            tarjan(e[i].to);
            low[x] = min(low[x], low[e[i].to]);
        } else if (!color[e[i].to])
            low[x] = min(low[x], dfn[e[i].to]);
    }
    if (low[x] == dfn[x]) {
        color[x] = ++num;
        tot[num]++;
        while (st[top] != x) {
            color[st[top]] = num;
            tot[num]++;
            --top;
        }
        --top;
    }
}
int d[MAX_N];
int main () {
    memset(h, -1, sizeof(h));
    n = gi(), m = gi();
    for (register int i = 1; i <= m; i++) {
        int u = gi(), v = gi();
        Add_Edge(u, v);
    }
    for (register int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);
    int ans = 0, F = 0;
    for (register int x = 1; x <= n; x++)
        for (int i = h[x]; ~i; i = e[i].next)
        if (color[x] != color[e[i].to]) d[color[x]]++;
    for (register int i = 1; i <= num; i++) {
    	if (d[i] == 0) F++, ans = tot[i];
    }
    if (F == 1) printf("%d\n", ans);
	else printf("0\n");
    return 0;
}