题干：
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
题目大意：
给你一个N*M的矩阵，' . ' 表示土地，' m '表示一个人，' H '代表一个屋子，现在告诉你屋子数和人数相同，要让所有人进入屋子内，且一个屋子只能装一个人，人移动一格花费是1，且只能上下左右移动，那么问你总的最小花费是多少。
解题报告：
人连到起点，房子连到终点，每一个格子可以上下左右四周走，求一个二分图最优匹配就行了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
#include <iomanip>
using namespace std;
const int MAX = 2e5 + 5;
const int INF = 0x3f3f3f3f;
struct node {int to,c,w,ne;
} e[50005<<2];
int n,m,N;
int head[MAX],d[MAX],vis[MAX],tot,p[MAX];
void add(int u,int v,int c,int cost=0) {e[++tot].to = v;e[tot].c = c;e[tot].w = cost;e[tot].ne = head[u];head[u] = tot;e[++tot].to = u;e[tot].c = 0; e[tot].w = -cost;e[tot].ne = head[v];head[v] = tot;
}
bool bfs(int s,int t) {for(int i = 0; i<=N; i++) d[i]=INF,vis[i]=0;d[s]=0;queue<int>q;q.push(s);while(!q.empty()) {int u=q.front();q.pop();vis[u]=0;for(int i=head[u]; ~i; i=e[i].ne) {int j=e[i].to;if(e[i].c&&d[j]>d[u]+e[i].w) {d[j]=d[u]+e[i].w;p[j]=i;if(!vis[j])vis[j]=1,q.push(j);}}}return d[t]<INF;
}
int MCMF(int s,int t,int &flow) {ll ans=0;while(bfs(s,t)) {int x=t,f=INF;while(x!=s) {f = min(f,e[p[x]].c),x=e[p[x]^1].to;}flow += f;ans+=1LL*d[t]*f;x=t;while(x!=s) {e[p[x]].c-=f,e[p[x]^1].c+=f;x=e[p[x]^1].to;}}return ans;
}
void init(int NN) {tot=1;for(int i = 0; i<=NN; i++) head[i] = -1;
}
char s[105][105];
int id(int i,int j) {return (i-1)*m+j;
}
int nx[4] = {0,1,0,-1};
int ny[4] = {1,0,-1,0};
int main()
{while(~scanf("%d%d",&n,&m)) {if(n+m==0) break;int st=0,ed=n*m+1;N=n*m+1; init(n*m+1);for(int i = 1; i<=n; i++) {scanf("%s",s[i]+1);for(int j = 1; j<=m; j++) {if(s[i][j] == 'm') add(st,id(i,j),1,0);if(s[i][j] == 'H') add(id(i,j),ed,1,0); }}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {for(int k = 0; k<4; k++) {int tx = i+nx[k];int ty = j+ny[k];if(tx<1||tx>n||ty<1||ty>m) continue;add(id(i,j),id(tx,ty),INF,1);}}}int ansf=0;int ansc=MCMF(st,ed,ansf);printf("%d\n",ansc); }return 0 ;
}
