POJ 3281 Dining

题目链接

题意：n个牛，每个牛有一些喜欢的食物和饮料，每种食物饮料只有一个，问最大能匹配上多少只牛每个牛都能吃上喜欢的食物和喜欢的饮料

思路：最大流，建模源点到每个食物连一条边，容量为1，每个饮料向汇点连一条边容量为1，然后由于每个牛有容量1，所以把牛进行拆点，然后食物连向牛的入点，牛的出点连向食物，跑一下最大流即可

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 505;
const int MAXEDGE = 100005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type cap, flow;
	Edge() {}
	Edge(int u, int v, Type cap, Type flow) {
		this->u = u;
		this->v = v;
		this->cap = cap;
		this->flow = flow;
	}
};

struct Dinic {
	int n, m, s, t;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	bool vis[MAXNODE];
	Type d[MAXNODE];
	int cur[MAXNODE];
	vector<int> cut;

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}
	void add_Edge(int u, int v, Type cap) {
		edges[m] = Edge(u, v, cap, 0);
		next[m] = first[u];
		first[u] = m++;
		edges[m] = Edge(v, u, 0, 0);
		next[m] = first[v];
		first[v] = m++;
	}

	bool bfs() {
		memset(vis, false, sizeof(vis));
		queue<int> Q;
		Q.push(s);
		d[s] = 0;
		vis[s] = true;
		while (!Q.empty()) {
			int u = Q.front(); Q.pop();
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (!vis[e.v] && e.cap > e.flow) {
					vis[e.v] = true;
					d[e.v] = d[u] + 1;
					Q.push(e.v);
				}
			}
		}
		return vis[t];
	}

	Type dfs(int u, Type a) {
		if (u == t || a == 0) return a;
		Type flow = 0, f;
		for (int &i = cur[u]; i != -1; i = next[i]) {
			Edge& e = edges[i];
			if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
				e.flow += f;
				edges[i^1].flow -= f;
				flow += f;
				a -= f;
				if (a == 0) break;
			}
		}
		return flow;
	}

	Type Maxflow(int s, int t) {
		this->s = s; this->t = t;
		Type flow = 0;
		while (bfs()) {
			for (int i = 0; i < n; i++)
				cur[i] = first[i];
			flow += dfs(s, INF);
		}
		return flow;
	}

	void MinCut() {
		cut.clear();
		for (int i = 0; i < m; i += 2) {
			if (vis[edges[i].u] && !vis[edges[i].v])
				cut.push_back(i);
		}
	}
} gao;

int f, d, n;

int main() {
	while (~scanf("%d%d%d", &n, &f, &d)) {
		int s = 0, t = f + d + n * 2 + 1;
		gao.init(t + 1);
		for (int i = 1; i <= f; i++)
			gao.add_Edge(s, i, 1);
		for (int i = 1; i <= d; i++)
			gao.add_Edge(f + i, t, 1);
		for (int i = 1; i <= n; i++) {
			gao.add_Edge(i + f + d, n + f + d + i, 1);
			int fn, dn;
			scanf("%d%d", &fn, &dn);
			int v;
			while (fn--) {
				scanf("%d", &v);
				gao.add_Edge(v, i + f + d, 1);
			}
			while (dn--) {
				scanf("%d", &v);
				gao.add_Edge(n + f + d + i, f + v, 1);
			}
		}
		printf("%d\n", gao.Maxflow(s, t));
	}
	return 0;
}

POJ 3281 Dining(最大流）