N - Optimal Milking POJ - 2112（二分 + Floyd + 二分图多重匹配）

题目链接
解题思路：

• 先算出每头奶牛到每块挤奶地的最短距离，如果不存在则为inf
• 二分最长距离，并以此建边，进行多重匹配

AC代码：

// Test1.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//
//二分答案 + 二分图多重匹配 
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
const int maxn = 510;
int K, C, M, mmax;
int dis[maxn][maxn], graph[maxn][maxn], match[maxn][maxn], vis[maxn];
void Floyd() {
	for (int k = 1; k <= K + C; k++)
		for (int i = 1; i <= K + C; i++)
			for (int j = 1; j <= K + C; j++)
				if (i != j && dis[i][j] > dis[i][k] + dis[k][j])
					dis[i][j] = dis[i][k] + dis[k][j];
	for (int i = 1; i <= K + C; i++)
		for (int j = 1; j <= K + C; j++)
			mmax = max(dis[i][j], mmax);
}
void Init() {
	memset(dis, 0x3f, sizeof(dis));
	int x;
	for (int i = 1; i <= K + C; i++)
		for (int j = 1; j <= K + C; j++) {
			cin >> x;
			if (x) dis[i][j] = x;
		}
	Floyd();
}
void Build_graph(int mid) {
	for (int i = K + 1; i <= K + C; i++)
		for (int j = 1; j <= K; j++)
			if (dis[i][j] <= mid) graph[i][j] = 1;
			else graph[i][j] = 0;
}
bool dfs(int x, int mid) {
	for (int i = 1; i <= K; i++) {
		if (graph[x][i] && !vis[i]) {
			vis[i] = 1;
			if (match[i][0] < M) {
				match[i][++match[i][0]] = x;
				return true;
			}
			for (int j = 1; j <= match[i][0]; j++)
				if (dfs(match[i][j], mid)) {
					match[i][j] = x;
					return true;
				}
		}
	}
	return false;
}
int Maxmatch(int mid) {
	int sum = 0;
	memset(match, 0, sizeof(match));
	for (int i = K + 1; i <= K + C; i++) {
		memset(vis, 0, sizeof(vis));
		if (dfs(i, mid)) sum++;
	}
	return sum;
}
int main() {
	cin >> K >> C >> M;
	Init();
	int lef = 1, rig = mmax, mid = (lef + rig) >> 1, ans;
	while (lef <= rig) {
		Build_graph(mid);
		if (Maxmatch(mid) == C) rig = mid - 1, ans = mid;
		else lef = mid + 1;
		mid = (lef + rig) >> 1;
	}
	cout << ans << endl;
	return 0;
}

本文地址：https://blog.csdn.net/weixin_45691711/article/details/107284547