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HDU 2426 Interesting Housing Problem(KM完美匹配)

2018年10月11日 ⁄ 综合 ⁄ 共 1481字 ⁄ 字号 评论关闭

HDU 2426 Interesting Housing Problem

题目链接

题意:n个学生,m个房间,给定一些学生想住房的喜欢度,找一个最优方案使得每个学生分配一个房间,并且使得喜欢度最大,注意这题有个坑,就是学生不会住喜欢度为负的房间

思路:很明显的KM最大匹配问题,喜欢度为负直接不连边即可

代码:

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

const int MAXNODE = 505;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct KM {
	int n;
	Type g[MAXNODE][MAXNODE];
	Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];
	int left[MAXNODE], right[MAXNODE];
	bool S[MAXNODE], T[MAXNODE];

	void init(int n) {
		this->n = n;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				g[i][j] = -INF;
	}

	void add_Edge(int u, int v, Type val) {
		g[u][v] = max(g[u][v], val);
	}

	bool dfs(int i) {
		S[i] = true;
		for (int j = 0; j < n; j++) {
			if (T[j]) continue;
			Type tmp = Lx[i] + Ly[j] - g[i][j];
			if (!tmp) {
				T[j] = true;
				if (left[j] == -1 || dfs(left[j])) {
					left[j] = i;
					right[i] = j;
					return true;
				}
			} else slack[j] = min(slack[j], tmp);
		}
		return false;
	}

	void update() {
		Type a = INF;
		for (int i = 0; i < n; i++)
			if (!T[i]) a = min(a, slack[i]);
		for (int i = 0; i < n; i++) {
			if (S[i]) Lx[i] -= a;
			if (T[i]) Ly[i] += a;
		}
	}

	int km(int tot) {
		for (int i = 0; i < n; i++) {
			left[i] = -1; right[i] = -1;
			Lx[i] = -INF; Ly[i] = 0;
			for (int j = 0; j < n; j++)
				Lx[i] = max(Lx[i], g[i][j]);
		}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) slack[j] = INF;
			while (1) {
				for (int j = 0; j < n; j++) S[j] = T[j] = false;
				if (dfs(i)) break;
				else update();
			}
		}
		int ans = 0;
		for (int i = 0; i < tot; i++) {
			if (right[i] == -1) return -1;
			if (g[i][right[i]] == -1) return -1;
			ans += g[i][right[i]];
		}
		return ans;
	}
} gao;

int n, m, k;

int main() {
	int cas = 0;
	while (~scanf("%d%d%d", &n, &m, &k)) {
		gao.init(max(n, m));
		int u, v, w;
		while (k--) {
			scanf("%d%d%d", &u, &v, &w);
			if (w >= 0) gao.add_Edge(u, v, w);
		}
		printf("Case %d: %d\n", ++cas, gao.km(n));
	}
	return 0;
}

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