HDU 3277 Marriage Match III

题目链接

题意：n个女孩n个男孩，每个女孩可以和一些男孩配对，此外还可以和k个任意的男孩配对，然后有些女孩是朋友，满足这个朋友圈里面的人，如果有一个能和某个男孩配对，其他就都可以，然后每轮要求每个女孩匹配到一个男孩，且每轮匹配到的都不同，问最多能匹配几轮

思路，比HDU3081多了一个条件，此外可以和k个任意的男孩配对，转化为模型，就是多了一个结点，有两种两边的方式，一种连向可以配对的，一种连向不能配对的，此外还要保证流量算在一起，这要怎么搞呢。
其实拆点就可以了，一个女孩拆成两个点，一个连可以配，一个连不能配，源点连向其中一点，容量为mid，然后两点之间，在连一条边，连接起来，这样就能保证总流量不会超过mid了，其他都和上一题差不不多，不过这题数据比上一题大，要注意一开是预处理关系的做法

代码：

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

const int MAXNODE = 805;
const int MAXEDGE = 200005;

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;

const int N = 205;

int t, n, m, k, f, g[N][N], parent[N];
int u[N * N], v[N * N];

int find(int x) {
	return x == parent[x] ? x : parent[x] = find(parent[x]);
}

bool judge(int mid) {
	int s = 0, t = n * 3 + 1;
	gao.init(3 * n + 2);
	for (int i = 1; i <= n; i++) {
		gao.add_Edge(s, i, mid);
		gao.add_Edge(i, i + n, k);
		gao.add_Edge(i + 2 * n, t, mid);
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (g[i][j]) gao.add_Edge(i, j + 2 * n, 1);
			else gao.add_Edge(i + n, j + 2 * n, 1);
		}
	}
	return gao.Maxflow(s, t) == n * mid;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d%d%d", &n, &m, &k, &f);
		memset(g, 0, sizeof(g));
		for (int i = 1; i <= n; i++) parent[i] = i;
		for (int i = 0; i < m; i++)
			scanf("%d%d", &u[i], &v[i]);
		int a, b;
		while (f--) {
			scanf("%d%d", &a, &b);
			int pa = find(a);
			int pb = find(b);
			if (pa != pb) parent[pa] = pb;
		}
		for (int i = 0; i < m; i++)
			g[find(u[i])][v[i]] = 1;
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				g[i][j] |= g[find(i)][j];
		int l = 1, r = n + 1;
		while (l < r) {
			int mid = (l + r) / 2;
			if (judge(mid)) l = mid + 1;
			else r = mid;
		}
		printf("%d\n", l - 1);
	}
	return 0;
}

HDU 3277 Marriage Match III（二分+最大流）