Going Home

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

思路：提取出各个点，然后建边，求最大权匹配。

#include"stdio.h"
#include"string.h"
#include"math.h"
#define N 105
#define mmax(a,b) ((a)>(b)?(a):(b))
#define mmin(a,b) ((a)<(b)?(a):(b))
const int inf=0x7fffffff;
int g[N][N],slack[N];
int link[N],lx[N],ly[N];
int visx[N],visy[N],n;
struct node
{
    int x,y;
}h[N],man[N];
int find(int k,int n)
{
    visx[k]=1;
    int i,d;
    for(i=0;i<n;i++)
    {
        if(visy[i])
            continue;
        d=lx[k]+ly[i]-g[k][i];
        if(d==0)
        {
            visy[i]=1;
            if(link[i]==-1||find(link[i],n))
            {
                link[i]=k;
                return 1;
            }
        }
        else
            slack[i]=mmin(slack[i],d);
    }
    return 0;
}
int KM(int n)
{
    int i,j;
    memset(lx,0,sizeof(lx));
    memset(ly,0,sizeof(ly));
    memset(link,-1,sizeof(link));
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            lx[i]=mmax(lx[i],g[i][j]);
        }
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            slack[j]=inf;
        }
        while(1)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(find(i,n))
                break;
            else
            {
                int d=inf;
                for(j=0;j<n;j++)
                {
                    if(!visy[j])
                    {
                        d=mmin(d,slack[j]);
                    }
                }
                for(j=0;j<n;j++)
                {
                    if(visx[j])
                        lx[j]-=d;
                    if(visy[j])
                        ly[j]+=d;
                }
            }
        }
    }
    int ans=0;
    for(i=0;i<n;i++)
    {
        ans+=g[link[i]][i];
    }
    return ans;
}
int main()
{
    int i,j,n,m;
    char str[N][N];
    while(scanf("%d%d",&n,&m),n||m)
    {
        int n1,n2;
        n1=n2=0;
        for(i=0;i<n;i++)
        {
            scanf("%s",str[i]);
            for(j=0;j<m;j++)
            {
                if(str[i][j]=='H')
                {
                    h[n1].x=i;
                    h[n1++].y=j;
                }
                if(str[i][j]=='m')
                {
                    man[n2].x=i;
                    man[n2++].y=j;
                }
            }
        }
        for(i=0;i<n1;i++)
        {
            for(j=0;j<n2;j++)   //求出两点间的曼哈顿距离，即建图
            {
                g[i][j]=-abs(h[i].x-man[j].x)-abs(h[i].y-man[j].y);
            }
        }
        printf("%d\n",-KM(n1));
    }
    return 0;
}