HDU 3861 The King’s Problem

题目链接

题意：给定一个有向图，求最少划分成几个部分满足下面条件

互相可达的点必须分到一个集合
一个对点(u, v)必须至少有u可达v或者v可达u
一个点只能分到一个集合

思路：先强连通缩点，然后二分图匹配求最小路径覆盖

代码：

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

const int N = 5005;

int t, n, m;
vector<int> g[N];
stack<int> S;

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

void dfs(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(v);
			dfn[u] = min(dfn[u], dfn[v]);
		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
	}
	if (dfn[u] == pre[u]) {
		sccn++;
		while (1) {
			int x = S.top(); S.pop();
			sccno[x] = sccn;
			if (x == u) break;
		}
	}
}

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

int left[N], vis[N];
vector<int> scc[N];

bool match(int u) {
	for (int i = 0; i < scc[u].size(); i++) {
		int v = scc[u][i];
		if (vis[v]) continue;
		vis[v] = 1;
		if (!left[v] || match(left[v])) {
			left[v] = u;
			return true;
		}
	}
	return false;
}

int hungary() {
	memset(left, 0, sizeof(left));
	int ans = 0;
	for (int i = 1; i <= sccn; i++) {
		memset(vis, 0, sizeof(vis));
		if (match(i)) ans++;
	}
	return sccn - ans;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) g[i].clear();
		int u, v;
		while (m--) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
		}
		find_scc();
		for (int i = 1; i <= sccn; i++) scc[i].clear();
		for (int u = 1; u <= n; u++) {
			for (int j = 0; j < g[u].size(); j++) {
				int v = g[u][j];
				if (sccno[u] == sccno[v]) continue;
				scc[sccno[u]].push_back(sccno[v]);
			}
		}
		printf("%d\n", hungary());
	}
	return 0;
}

HDU 3861 The King’s Problem（强连通+二分图最小路径覆盖）