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. 
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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
消停套模板，没啥好说de 
#include <stdio.h>
#include <string.h>
#define M 310
#define inf 0x3f3f3f3f

int abs(int x)
{
    return x>0?x:-x;
}
int n,m,nx,ny;
int link[M],lx[M],ly[M],slack[M];///lx,ly为顶标，nx,ny分别为x点集y点集的个数
int visx[M],visy[M],w[M][M];

int DFS(int x)
{
    visx[x] = 1;
    for (int y = 1; y <= ny; y ++)
    {
        if (visy[y]) continue;
        int t = lx[x] + ly[y] - w[x][y];
        if (t == 0)
        {
            visy[y] = 1;
            if (link[y] == -1||DFS(link[y]))
            {
                link[y] = x;
                return 1;
            }
        }
        else if (slack[y] > t)  ///不在相等子图中slack 取最小的
            slack[y] = t;
    }
    return 0;
}
int KM()
{
    int i,j;
    memset (link,-1,sizeof(link));
    memset (ly,0,sizeof(ly));
    for (i = 1; i <= nx; i ++)          ///lx初始化为与它关联边中最大的
        for (j = 1,lx[i] = -inf; j <= ny; j ++)
            if (w[i][j] > lx[i])
                lx[i] = w[i][j];

    for (int x = 1; x <= nx; x ++)
    {
        for (i = 1; i <= ny; i ++)
            slack[i] = inf;
        while (1)
        {
            memset (visx,0,sizeof(visx));
            memset (visy,0,sizeof(visy));
            if (DFS(x))     ///若成功（找到了增广轨），则该点增广完成，进入下一个点的增广
                break;  ///若失败（没有找到增广轨），则需要改变一些点的标号，使得图中可行边的数量增加。
            ///方法为：将所有在增广轨中（就是在增广过程中遍历到）的X方点的标号全部减去一个常数d，
            ///所有在增广轨中的Y方点的标号全部加上一个常数d
            int d = inf;
            for (i = 1; i <= ny; i ++)
                if (!visy[i]&&d > slack[i])
                    d = slack[i];
            for (i = 1; i <= nx; i ++)
                if (visx[i])
                    lx[i] -= d;
            for (i = 1; i <= ny; i ++) ///修改顶标后，要把所有不在交错树中的Y顶点的slack值都减去d
                if (visy[i])
                    ly[i] += d;
                else
                    slack[i] -= d;
        }
    }
    int res = 0;
    for (i = 1; i <= ny; i ++)
        if (link[i] > -1)
            res += w[link[i]][i];
    return res;
}

char a[104][104];
struct note
{
    int x,y;
}node1[105],node2[105];

int main()
{
    //freopen("cin.txt","r",stdin);
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        nx=ny=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i]);
            for(int j=0;j<m;j++)
            {
                if(a[i][j]=='m')  node1[++nx].x=i,node1[nx].y=j;
                if(a[i][j]=='H')  node2[++ny].x=i,node2[ny].y=j;
            }
        }
        for(int i=1;i<=nx;i++)
        {
            for(int j=1;j<=ny;j++)
            {
                w[i][j]=-abs(node1[i].x-node2[j].x)-abs(node1[i].y-node2[j].y);
            }
        }
        printf("%d\n",-KM());
    }
    return 0;
}