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.mH.5 5HH..m...............mm..H7 8...H.......H.......H....mmmHmmmm...H.......H.......H....0 0

Sample Output

21028

比较裸的费用流，可以当做模板用，建图方式在代码中有

#include <cstdio>#include <cstring>#include <queue>#include <math.h>#include <algorithm>using namespace std;#define maxn 310#define INF 0x3f3f3f3fstruct node{    int v , w , s ;    int next ;} p[maxn*100];int head[maxn] , cnt , vis[maxn] , pre[maxn] , dis[maxn] ;queue <int> q ;struct n{    int x , y ;} mm[120] , hh[120] ; //mm储存人的坐标，hh存储房屋的坐标char str[120][120] ;void add(int u,int v,int w,int s){    p[cnt].v = v ; p[cnt].w = w ; p[cnt].s = s ;    p[cnt].next = head[u] ; head[u] = cnt++ ;    p[cnt].v = u ; p[cnt].w = 0 ; p[cnt].s = -s ;    p[cnt].next = head[v] ; head[v] = cnt++ ;}int spfa(int s,int t){    int u , v , i ;    memset(pre,-1,sizeof(pre));    memset(dis,INF,sizeof(dis));    memset(vis,0,sizeof(vis));    while( !q.empty() )        q.pop();    q.push(s) ;    vis[s] = 1 ;    dis[s] = 0 ;    while( !q.empty() )    {        u = q.front();        q.pop();        vis[u] = 0 ;        for(i = head[u] ; i != -1 ; i = p[i].next)        {            v = p[i].v ;            if( p[i].w && dis[v] > dis[u] + p[i].s )            {                dis[v] = dis[u] + p[i].s ;                pre[v] = i ;                if( !vis[v] )                {                    vis[v] = 1 ;                    q.push(v) ;                }            }        }    }    if( pre[t] == -1 )        return 0 ;    return 1 ;}void f(int s,int t){    int i , min1 , ans = 0 ;    while( spfa(s,t) )    {        min1 = INF ;        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])            if( p[i].w < min1 )                min1 = p[i].w ;        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])        {            p[i].w -= min1 ;            p[i^1].w += min1 ;            ans += p[i].s*min1 ;        }    }    printf("%d\n", ans);}int main(){    int n , nh , nm , m , i , j ;    while(scanf("%d %d", &n, &m)!=EOF)    {        /*建图方式，源点连接到人（容量是1，花费是0），人连接到所有的房屋（容量是1，花费是人与房屋的距离），房屋连接到汇点（容量是1，花费是0），其他的就是模板了*/        cnt = 0 ;        nh = 1 ;        nm = 1 ;        memset(head,-1,sizeof(head));        if(n == 0 && m == 0)            break;        for(i = 0 ; i < n ; i++)            scanf("%s", str[i]);        for(i = 0 ; i < n ; i++)            for(j = 0 ; j < m ; j++)                if( str[i][j] == 'm' )                {                    mm[nm].x = i ;                    mm[nm++].y = j ;                }                else if( str[i][j] == 'H' )                {                    hh[nh].x = i ;                    hh[nh++].y = j ;                }        for(i = 1 ; i < nm ; i++)            add(0,i,1,0);        for(i = 1 ; i < nm ; i++)            for(j = 1 ; j < nh ; j++)            {                add(i,j+100,1, fabs(mm[i].x-hh[j].x)+fabs(mm[i].y-hh[j].y) );//由人和房屋的坐标计算人到房屋的花费            }        for(i = 1 ; i < nh ; i++)            add(i+100,201,1,0);//最多人或房屋均为100，设人的标号为1到100，房屋的标号101到200，源点0，汇点201        f(0,201);    }    return 0;}