洛谷 P2341 【HAOI2006】受歡迎的牛

題目鏈接

https://www.luogu.org/problem/P2341

分析

Tarjan縮點，最後DAG中，唯一的出度爲 $0$ 的點的大小即爲答案。

AC代碼

#include <cstdio>
#include <algorithm>
#include <stack>

using namespace std;

inline int read() {
	int num = 0;
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	while (c >= '0' && c <= '9')
		num = num * 10 + c - '0', c = getchar();
	return num;
}

const int maxn = 1e4 + 5, maxm = 5e4 + 5;

int head[maxn], eid;

struct Edge {
	int v, next;
} edge[maxm];

inline void insert(int u, int v) {
	edge[++eid].v = v;
	edge[eid].next = head[u];
	head[u] = eid;
}

int dfn[maxn], low[maxn], tot, scc[maxn], num[maxn], cnt, out[maxn], ans;
stack<int> s;

void tarjan(int u) {
	dfn[u] = low[u] = ++tot;
	s.push(u);
	for (int p = head[u]; p; p = edge[p].next) {
		int v = edge[p].v;
		if (!dfn[v]) tarjan(v), low[u] = min(low[u], low[v]);
		else low[u] = min(low[u], dfn[v]);
	}
	if (low[u] == dfn[u]) {
		++cnt;
		while (s.top() != u) scc[s.top()] = cnt, ++num[cnt], s.pop();
		scc[u] = cnt, ++num[cnt], s.pop();
	}
}

int main() {
	int n = read(), m = read();
	for (int i = 1; i <= m; ++i) {
		int u = read(), v = read();
		insert(u, v);
	}
	for (int i = 1; i <= n; ++i)
		if (!dfn[i]) tarjan(i);
	for (int u = 1; u <= n; ++u)
		for (int p = head[u]; p; p = edge[p].next) {
			int v = edge[p].v;
			if (scc[u] != scc[v]) ++out[scc[u]];
		}
	for (int i = 1; i <= cnt; ++i)
		if (!out[i]) {
			if (!ans) ans = num[i];
			else {
				ans = 0;
				break;
			}
		}
	printf("%d", ans);
	return 0;
}