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

Output

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

KM算法！

其实这个题求的是最小权匹配，，但有些题目最小不一定好求，于是我们可以换一种思维，将有所的距离变成负的，那么我们要求的就是最大权匹配！（在一位大牛的指点下）
废话不多说，看程序：
  1 #include<iostream>
  2 using namespace std;
  3 #define Inf 10000000;
  4 int map[105][105];
  5 int slack[105];
  6 int lx[105],ly[105];
  7 bool x[105],y[105];
  8 int link[105];
  9 int n,m;
 10 char mm[105][105];
 11 bool dfs(int v)
 12 {
 13      x[v]=true;
 14      int t;
 15      for (int i=0;i<m;i++)
 16      {
 17          if (!y[i])
 18          {
 19             t=lx[v]+ly[i]-map[v][i];
 20             if (t==0)
 21             {
 22                y[i]=true;
 23                if (link[i]==-1||dfs(link[i]))
 24                {
 25                   link[i]=v;
 26                   return true;
 27                }
 28             }
 29             else slack[i]=min(slack[i],t);
 30          }
 31      }
 32      return false;
 33 }
 34 void KM()
 35 {
 36      int i,j,k;
 37      memset(link,-1,sizeof(link));
 38      for (i=0;i<=m;i++)
 39      {
 40          lx[i]=-Inf;
 41          ly[i]=0;
 42      }
 43      for (i=0;i<m;i++)
 44      {
 45          while (1)
 46          {
 47                memset(x,0,sizeof(x));
 48                memset(y,0,sizeof(y));
 49                
 50                for (j=0;j<m;j++) slack[j]=Inf;
 51                    
 52                if (dfs(i))break;
 53                
 54                int d=Inf;
 55                for (j=0;j<m;j++)
 56                    if (!y[j])d=min(slack[j],d);
 57                    
 58                for (j=0;j<m;j++)
 59                {
 60                    if (x[j])lx[j]-=d;
 61                    if (y[j])ly[j]+=d;
 62                }
 63                for (j=0;j<m;j++)
 64                    if (!y[j])slack[j]-=d;
 65          }
 66      }
 67 }
 68 int main()
 69 {
 70     int i,j,k,t,l,v;
 71     while (cin>>k>>t)
 72     {
 73           if (k+t==0)break;
 74           for (i=1;i<=k;i++)
 75           for (j=1;j<=t;j++)
 76               scanf(" %c",&mm[i][j]);
 77           memset(map,0,sizeof(map));
 78           n=0,m=0;
 79           for (i=1;i<=k;i++)
 80           for (j=1;j<=t;j++)
 81           {
 82               if (mm[i][j]=='H')
 83               {
 84                  n=0;
 85                  for (l=1;l<=k;l++)
 86                  for (v=1;v<=t;v++)
 87                  {
 88                      if (mm[l][v]=='m')
 89                      {
 90                         map[m][n]=-(abs(i-l)+abs(j-v));
 91                         n++;
 92                      }
 93                  }
 94                  m++;
 95               }
 96           }
 97           KM();
 98           int sum=0;
 99           for (i=0;i<m;i++)
100               sum+=map[link[i]][i];
101           cout<<0-sum<<endl;
102     }
103 return 0;
104 }
105  
106