[洛谷P1402]酒店之王

题目大意：有$n$个人，$p$个房间，$q$道菜（$n,p,q\leqslant100$），每个人有自己喜欢的房间和菜，问最多使得多少个人高兴（即拥有喜欢的房间和菜）

题解：网络流，源点向每道菜连边，每道菜向喜欢的人连边，每个人向喜欢的房间连边，每个房间再向汇点连边。发现这样人会重复，于是把人拆点即可。

卡点：无

C++ Code：

#include <algorithm>
#include <cstdio>
#include <iostream>
#define maxn 111
const int N = maxn << 2, M = maxn * maxn * 2 + maxn * 3;
const int inf = 0x3f3f3f3f;

namespace Network_Flow {
	int lst[N], head[N], cnt = 1;
	struct Edge {
		int to, nxt, w;
	} e[M << 1];
	inline void addedge(int a, int b, int c = 1) {
		e[++cnt] = (Edge) { b, head[a], c }; head[a] = cnt;
		e[++cnt] = (Edge) { a, head[b], 0 }; head[b] = cnt;
	}

	int n, st, ed, MF;
	int GAP[N], d[N];
	int q[N], h, t;

	void init() {
		GAP[d[ed] = 1] = 1;
		for (int i = st; i <= ed; ++i) lst[i] = head[i];
		q[h = t = 0] = ed;
		while (h <= t) {
			int u = q[h++];
			for (int i = head[u]; i; i = e[i].nxt) {
				int v = e[i].to;
				if (!d[v]) {
					d[v] = d[u] + 1;
					++GAP[d[v]];
					q[++t] = v;
				}
			}
		}
	}
	int dfs(int u, int low) {
		if (!low || u == ed) return low;
		int w, res = 0;
		for (int &i = lst[u]; i; i = e[i].nxt) if (e[i].w) {
			int v = e[i].to;
			if (d[u] == d[v] + 1) {
				w = dfs(v, std::min(low, e[i].w));
				res += w, low -= w;
				e[i].w -= w, e[i ^ 1].w += w;
				if (!low) return res;
			}
		}
		if (!(--GAP[d[u]])) d[st] = n + 1;
		++GAP[++d[u]], lst[u] = head[u];
		return res;
	}
	void ISAP(int S, int T) {
		st = S, ed = T;
		n = ed - st + 1;
		init();
		while (d[st] <= n) MF += dfs(st, inf);
	}
}
using Network_Flow::addedge;

int n, p, q;
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> n >> p >> q;
	int st = 0, ed = p + q + n + n + 1;
	for (int i = 1; i <= p; ++i) addedge(st, i);
	for (int i = 1, t; i <= n; ++i) {
		for (int j = 1; j <= p; ++j) {
			std::cin >> t;
			if (t) addedge(j, i + p);
		}
	}
	for (int i = 1; i <= n; ++i) addedge(i + p, i + n + p);
	for (int i = 1, t; i <= n; ++i) {
		for (int j = 1; j <= q; ++j) {
			std::cin >> t;
			if (t) addedge(i + n + p, j + n + n + p);
		}
	}
	for (int i = 1; i <= p; ++i) addedge(i + n + n + p, ed);
	Network_Flow::ISAP(st, ed);
	std::cout << Network_Flow::MF << std::endl;
	return 0;
}

标签：连边,洛谷,每道菜,P1402,房间,之王,int,maxn,include
来源： https://www.cnblogs.com/Memory-of-winter/p/10342645.html