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.

2 2

.m

H.

5 5

HH..m

.....

.....

.....

mm..H

7 8

...H....

...H....

...H....

mmmHmmmm

...H....

...H....

...H....

0 0

2

10

28

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <cmath>

 using namespace std;

 typedef long long ll;

 const int MAXX=;

 const int INF=0x3f3f3f3f;

 struct node

 {

     int st;

     int to;

     int next;

     int cap;

     int cost;

 }edge[MAXX];

 char mp[][];

 int head[MAXX],tol;

 int pre[MAXX],dis[MAXX];

 bool vis[MAXX];

 int n,m,p;

 struct node1

 {

     int x,y;

     node1(){}

     node1(int a,int b)

     {

         x=a;

         y=b;

     }

 };

 vector<node1> v1,v2;

 void init()

 {

     tol=;

     memset(head,-,sizeof(head));

     v1.clear();

     v2.clear();

 }

 void addedge(int u,int v,int cap,int cost)

 {

     edge[tol].st=u;

     edge[tol].to=v;

     edge[tol].cap=cap;

     edge[tol].cost=cost;

     edge[tol].next=head[u];

     head[u]=tol++;

     edge[tol].st=v;

     edge[tol].to=u;

     edge[tol].cap=;

     edge[tol].cost=-cost;

     edge[tol].next=head[v];

     head[v]=tol++;

 }

 bool SPFA(int s,int t)

 {

     queue<int> q;

     memset(dis,INF,sizeof(dis));

     memset(vis,,sizeof(vis));

     memset(pre,-,sizeof(pre));

     dis[s]=;

     vis[s]=;

     q.push(s);

     while(!q.empty())

     {

         int u=q.front(); q.pop();

         vis[u]=;

         for(int i=head[u];i!=-;i=edge[i].next)

         {

             int to=edge[i].to;

             if(edge[i].cap>&&dis[to]>dis[u]+edge[i].cost)

             {

                 dis[to]=dis[u]+edge[i].cost;

                 pre[to]=i;

                 if(!vis[to])

                 {

                     vis[to]=;

                     q.push(to);

                 }

             }

         }

     }

     if(pre[t]==-)return ;

     return ;

 }

 int minCostMaxFlow(int s,int t)

 {

     int cost=;

     while(SPFA(s,t))

     {

         int minn=INF;

         for(int i=pre[t];i!=-;i=pre[edge[i].st])

             minn=min(minn,edge[i].cap);

         for(int i=pre[t];i!=-;i=pre[edge[i].st])

         {

             edge[i].cap-=minn;

             edge[i^].cap+=minn;

         }

         cost+=minn*dis[t];

     }

     return cost;

 }

 int main()

 {

     while(scanf("%d%d",&n,&m)&&m&&n)

     {

         getchar();

         init();

         for(int i=;i<n;i++)

         {

             scanf("%s",mp[i]);

             getchar();

         }

         for(int i=;i<n;i++)

         for(int j=;j<m;j++)

         {

             if(mp[i][j]=='m')

                 v1.push_back(node1(i,j));

             if(mp[i][j]=='H')

                 v2.push_back(node1(i,j));

         }

         int l=v1.size();

         int r=v2.size();

         for(int i=;i<v1.size();i++)

         {

             node1 N1=v1[i];

             int x=i+;

             addedge(,x,,);

             for(int j=;j<v2.size();j++)

             {

                 node1 N2=v2[j];

                 int y=j+l+;

                 int D=abs(N1.x-N2.x)+abs(N1.y-N2.y);

                 addedge(x,y,,D);

                 addedge(y,x,,D);

                 if(i==l-)

                 addedge(y,l+r+,,);

             }

         }

         int ans=minCostMaxFlow(,l++r);

         printf("%d\n",ans);

     }

     return ;

 }