hdu 1533 Going Home(最小费用最大流)

题目链接：

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题目大意：

给出一张地图，给出人和屋子的位置，问每个人都有屋子住的走的最小的距离之和

题目分析：

首先屋子的数量和人的数量相等，那么每个人一定有屋子住，要求最小的花费，花费就是人到屋子的距离，所以就是一个最优匹配的模板题，当然也可以看做是最小费用最大流的模板题，建图方法也很朴素，就是源点连向所有的人连边，容量为1，花费为0，汇点向所有的屋子连边，容量为1，花费为0，然后每个人到每个屋子的距离作为花费，1作为容量，在人和屋子之间建边

代码如下:

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
#define MAX 105
#define INF 0x7fffffff

using namespace std;

int flow[MAX<<1][MAX<<1];
int cost[MAX<<1][MAX<<1];
int map[MAX][MAX];
int pre[MAX<<1],dis[MAX<<1];
int minflow[MAX<<1],mark[MAX<<1];
char str[MAX][MAX];

struct Node
{
    int x,y;
}house[MAX],man[MAX];

int spfa ( int start , int end )
{
    queue<int> q;
    //memset ( dis , 0x3f , sizeof ( dis ));
    memset ( mark , 0 , sizeof ( mark ));
    memset ( pre , -1 , sizeof ( pre ));
    //memset ( minflow , INF , sizeof ( minflow));
    for ( int i = 0 ; i < (MAX<<1) ; i++ )
        dis[i] = minflow[i] = INF;
    q.push ( start );
    dis[start] = 0;
    mark[start] = 1;
    while ( !q.empty())
    {
        int u = q.front();
        q.pop();
        mark[u] = 0;
        for ( int i = 0 ; i <= end ; i++ )
            if (flow[u][i]&& dis[i] > dis[u] + cost[u][i] )
            {
                dis[i] = dis[u] + cost[u][i];
                pre[i] = u;
                minflow[i] = min ( minflow[u] , flow[u][i] );
                if ( !mark[i] )
                {
                    mark[i] = 1;
                    q.push ( i );
                }
            }
    }
    return dis[end]!= INF;
}

int maxflow_mincost ( int start , int end )
{
    int i,x,ans = 0;
    while ( spfa(start,end) )
    {
        x = end;
        while ( pre[x] != -1 )
        {
            flow[pre[x]][x] -= minflow[end];
            flow[x][pre[x]] += minflow[end];
            x = pre[x];
        }
        ans += dis[end];
    }
    return ans;
}

int n,m,k1,k2,t;

int main ( )
{
    while (~scanf ( "%d%d" , &n , &m ) , n+m )
    {
        k1 = k2 = 1;
        memset ( flow , 0 , sizeof (flow ) );
        for ( int i = 0 ; i < n ; i++ )
        {
            scanf ( "%s" , str[i] );
            for ( int j = 0 ; j < m ; j++ )
            {
                if ( str[i][j] == 'H' )
                {
                    house[k1].x = i;
                    house[k1++].y = j;
                }
                else if ( str[i][j] == 'm')
                {
                    man[k2].x = i;
                    man[k2++].y = j;
                }
            }
        }

        n = k1-1;

        for ( int i= 0 ; i < 2*n+1 ; i++)
            for ( int j = 0 ; j < 2*n+1 ; j++ )
                cost[i][j] = INF;

        for ( int i = 1 ; i <= n ; i++ )
        {
            for ( int j = n+1 ; j < 2*n+1 ; j++ )
            {
                cost[i][j] = abs ( house[j-n].x - man[i].x )
                        +abs ( house[j-n].y - man[i].y );
                cost[j][i] = -cost[i][j];
                flow[i][j] = 1;
            }

            cost[i][0] =cost[0][i] = 0;
            flow[0][i] = 1;
            cost[i+n][2*n+1] = cost[2*n+1][i+n] = 0;
            flow[i+n][2*n+1] = 1;
        }

        printf ( "%d\n" , maxflow_mincost( 0 , 2*n+1 ) );
    }
}