# POJ 2195 Going Home （最小费用最大流）

Going Home

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22415 Accepted: 11330

Description

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a \$1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man.Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a ‘.’ means an empty space, an ‘H’ represents a house on that point, and am ‘m’ indicates there is a little man on that point.

You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

Input

There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of ‘H’s and ‘m’s on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.

Output

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.

Sample Input

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

Sample Output

2
10
28

### AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<map>

const int MAXX = 10005;
const int INF = 0x3f3f3f3f;

struct edge
{
int to;     //边终点
int next;   //下一个兄弟位置
int cap;    //容量
int flow;   //流量
int cost;   //费用
} edge[MAXX<<2];

int pre[MAXX],dis[MAXX];
int N;  //节点总个数

void init(int n)
{
N=n;
tol=0;
}

void addedge(int u,int v,int cap,int cost)  //同时增加原边与反向边
{
edge[tol].to=v;
edge[tol].cap=cap;
edge[tol].cost=cost;
edge[tol].flow=0;
edge[tol].to=u;
edge[tol].cap=0;
edge[tol].cost=-cost;
edge[tol].flow=0;
}

/*
* SPFA 算法判断是否存在s到t的通路
*/
bool spfa(int s,int t)
{
queue<int>q;
bool vis[MAXX];
for(int i=0; i<N; i++)
{
dis[i]=INF;
vis[i]=false;
pre[i]=-1;
}
dis[s]=0;
vis[s]=true;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=false;
{
int v=edge[i].to;
if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)    //如果可以松弛该点
{
dis[v]=dis[u]+edge[i].cost;
pre[v]=i;
if(!vis[v]) //如果该点不在队列中，入队
{
vis[v]=true;
q.push(v);
}
}
}
}
return (pre[t]!=-1);    //返回是否s到t是否有路径
}

/*
* int s 起点
* int t 终点
*/
int minCostMaxFlow(int s,int t,int &cost)
{
int flow=0;
while(spfa(s,t))
{
int minn=INF;   //当前路径可增广值
for(int i=pre[t]; i!=-1; i=pre[edge[i^1].to])   //因为建图时每增加一条边会同时加入它的反向边，因此i^1为找出与i刚好相反的部分
{
if(minn>edge[i].cap-edge[i].flow)
minn=edge[i].cap-edge[i].flow;
}
for(int i=pre[t]; i!=-1; i=pre[edge[i^1].to])   //修改图，计算花费
{
edge[i].flow+=minn;
edge[i^1].flow-=minn;
cost+=edge[i].cost*minn;
}
flow+=minn;
}
return flow;
}

struct node
{
int x;
int y;
node() {}
node(int x,int y)
{
this->x=x;
this->y=y;
}
};

inline int getlen(node a,node b)
{
return abs(a.x-b.x)+abs(a.y-b.y);
}

int main()
{
int n,m;
char str[105][105];
while(~scanf("%d%d%*c",&n,&m)&&(n||m))
{
node H[5005],M[5005];   //模拟栈
int th=0,tm=0;
for(int i=0; i<n; i++)
{
gets(str[i]);
for(int j=0; j<m; j++)
{
if(str[i][j]=='H')
H[th++]=node(i,j);  //房子
else if(str[i][j]=='m')
M[tm++]=node(i,j);  //小人
}
}
init(th+tm+2);
for(int i=0; i<th; i++)
{