POJ 3281 Dining

題目鏈接

題意：n個牛。每一個牛有一些喜歡的食物和飲料。每種食物飲料僅僅有一個。問最大能匹配上多少僅僅牛每一個牛都能吃上喜歡的食物和喜歡的飲料

思路：最大流。建模源點到每一個食物連一條邊，容量為1，每一個飲料向匯點連一條邊容量為1，然後因為每一個牛有容量1。所以把牛進行拆點。然後食物連向牛的入點，牛的出點連向食物，跑一下最大流就可以

代碼：

#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）