POJ2112 Optimal Milking 二分法+网络流

题目大意是：K台挤奶机器，C头牛，K不超过30，C不超过200，每台挤奶机器最多可以为M台牛工作，给出这些牛和机器之间，牛和牛之间，机器与机器之间的距离，在保证让最多的牛都有机器挤奶的情况下，给出其中最长的一头牛移动的距离的最小值。

首先用Floyd求出任意两点之间的最短距离，然后再用二分法限定最多的移动距离d，在求最大流时，搜索增广路的时候同时也判断距离有没有超过d就行了。

#include <stdlib.h>
#include <stdio.h>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <iostream>
#include <queue>
#include <list>
#include <algorithm>
#include <stack>
#include <map>

#include<iostream>
#include<cstdio>
using namespace std;

#define MAXN 250
#define MAXDIS 100000
int G[MAXN][MAXN];
int D[MAXN][MAXN];
int Flow[MAXN][MAXN];
int Flow1[MAXN][MAXN];
bool Visited[MAXN];

void Floyd(int n)
{
	for (int i = 0; i < n + 2; i++)
	{
		for (int j = 0; j < n + 2; j++)
		{
			D[i][j] = MAXDIS;
			if (G[i][j] > 0)
			{
				D[i][j] = G[i][j];
			}
			if (i == j)
			{
				D[i][j] = 0;
			}
		}
	}
	for (int k = 0; k < n;k++)
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n;j++)
			{
				D[i][j] = min(D[i][j], D[i][k] + D[k][j]);
			}
		}
	}
}

int dfs(int s,int t, int f,int n, int limit)
{
	if (s == t)
	{
		return f;
	}
	Visited[s] = true;
	for (int i = 0; i < n; i++)
	{
		if (!Visited[i] && Flow[s][i] > 0 && D[s][i] <= limit)
		{
			Visited[i] = true;
			int d = dfs(i, t, min(f, Flow[s][i]), n, limit);
			if (d > 0)
			{
				Flow[s][i] -= d;
				Flow[i][s] += d;
				return d;
			}
		}
	}
	return 0;
}

int MaxFlow(int s, int t,int n, int limit)
{
	memset(Visited, 0, sizeof(Visited));
	int f = 0;
	int d = 0;
	while ((d=dfs(s, t, 1, n, limit)) > 0)
	{
		memset(Visited, 0, sizeof(Visited));
		f += d;
	}
	return f;
}

int main()
{
#ifdef _DEBUG
	freopen("d:\\in.txt", "r", stdin);
#endif
	int K, C, M;
	scanf("%d %d %d", &K, &C, &M);
	memset(G, 0, sizeof(G));
	for (int i = 0; i < K + C; i++)
	{
		for (int j = 0; j < K + C;j++)
		{
			int value;
			scanf("%d", &value);
			G[i][j] = value;
		}
	}
	Floyd(K + C);
	for (int i = 0; i < K + C; i++)
	{
		D[i][K + C] = 0;
		D[K + C][i] = 0;
		D[i][K + C + 1] = 0;
		D[K + C + 1][i] = 0;
	}
	memset(Flow, 0, sizeof(Flow));
	for (int i = 0; i < K;i++)
	{
		Flow[i][K + C + 1] = M;
	}
	int maxflow = min(C, K * M);
	for (int i = K; i < K + C;i++)
	{
		Flow[K + C][i] = 1;
		for (int j = 0; j < K;j++)
		{
			Flow[i][j] = 1;
		}
	}
	int l = 0;
	int r = MAXDIS;
	memcpy(Flow1, Flow, sizeof(Flow));
	while (r - l > 1)
	{
		memcpy(Flow, Flow1, sizeof(Flow));
		int mid = (l + r) / 2;
		int f = MaxFlow(K + C, K + C + 1, K + C + 2, mid);
		if (f >= maxflow)
		{
			r = mid;
		}
		else
			l = mid;
	}
	printf("%d\n", r);
	return 0;
}

POJ2112 Optimal Milking 二分法+网络流