POJ 1904 King's Quest

题目链接

题意：n个男人，每个人都有一个喜欢的女人列表，现在给一个完美匹配，问所有完美匹配中，每个人可能娶到的女人列表

思路：强连通，建图，男的连一条边指向女，然后完美匹配的边女的指向男，然后求强连通，在同一个强连通分支并且是自己想娶的的就可能娶到

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <stack>
using namespace std;

const int N = 4005;
int n;
vector<int> g[N];
int ans[N], an;
stack<int> S;

int pre[N], dfn[N], dfs_clock, sccno[N], sccn;

void dfs_scc(int u) {
	pre[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (!pre[v]) {
			dfs_scc(v);
			dfn[u] = min(dfn[u], dfn[v]);
		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
	}
	if (pre[u] == dfn[u]) {
		sccn++;
		while (1) {
			int x = S.top(); S.pop();
			sccno[x] = sccn;
			if (x == u) break;
		}
	}
}

void find_scc() {
	dfs_clock = sccn = 0;
	memset(pre, 0, sizeof(pre));
	memset(sccno, 0, sizeof(sccno));
	for (int i = 1; i <= 2 * n; i++)
		if (!pre[i]) dfs_scc(i);
}

int main() {
	while (~scanf("%d", &n)) {
		for (int i = 1; i <= 2 * n; i++) g[i].clear();
		int a, b;
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a);
			while (a--) {
				scanf("%d", &b);
				g[i].push_back(b + n);
			}
		}
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a);
			g[a + n].push_back(i);
		}
		find_scc();
		for (int i = 1; i <= n; i++) {
			an = 0;
			for (int j = 0; j < g[i].size(); j++) {
				if (sccno[i] == sccno[g[i][j]])
					ans[an++] = g[i][j] - n;
			}
			sort(ans, ans + an);
			printf("%d", an);
			for (int j = 0; j < an; j++)
				printf(" %d", ans[j]);
			printf("\n");
		}
	}
	return 0;
}