hdoj--1533--Going Home（KM）

Going Home

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3787    Accepted Submission(s): 1944

Problem 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

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

 

Source

Pacific Northwest 2004

 

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n*m的图中，m是人在的位置，H是房子，一个房子住一个人，求使得所有人到房子中 最少距离，

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f
int lx[110],ly[110];
int map[110][110];
bool visx[110],visy[110];
int slock[110];
int match[110];
int nx,ny;
int N,M,m_cnt,h_cnt;
char str[110][110];
int dis(int x1,int y1,int x2,int y2)
{
	return abs(x1-x2)+abs(y1-y2);
}
struct node
{
	int x,y;
};
node m[110],h[110];
void getMap()
{
	memset(str,'\0',sizeof(str));
	m_cnt=h_cnt=0;
	for(int i=0;i<N;i++)
	{
		scanf("%s",str[i]);
		for(int j=0;j<M;j++)
		{
			if(str[i][j]=='m')
			{
				m_cnt++;
				m[m_cnt].x=i;
				m[m_cnt].y=j;
			}
			if(str[i][j]=='H')
			{
				h_cnt++;
				h[h_cnt].x=i;
				h[h_cnt].y=j;
			}
		}
	}
	int k=m_cnt;
	nx=ny=k;
	for(int i=1;i<=k;i++)
	{
		for(int j=1;j<=k;j++)
		{
			int d=dis(h[i].x,h[i].y,m[j].x,m[j].y);
			map[i][j]=-d;
		}
	}
}
int DFS(int x)
{
	visx[x]=true;
	for(int y=1;y<=ny;y++)
	{
		if(visy[y]) continue;
		int t=lx[x]+ly[y]-map[x][y];
		if(t==0)
		{
			visy[y]=true;
			if(match[y]==-1||DFS(match[y]))
			{
				match[y]=x;
				return 1;
			}
		}
		else if(t<slock[y])
		slock[y]=t;
	}
	return 0;
}
void KM()
{
	memset(match,-1,sizeof(match));
	memset(ly,0,sizeof(ly));
	for(int x=1;x<=nx;x++)
	{
		lx[x]=-INF;
		for(int y=1;y<=ny;y++)
		lx[x]=max(lx[x],map[x][y]);
	}
	for(int x=1;x<=nx;x++)
	{
		for(int y=1;y<=ny;y++)
		slock[y]=INF;
		while(1)
		{
			memset(visx,false,sizeof(visx));
			memset(visy,false,sizeof(visy));
			if(DFS(x)) break;
			int d=INF;
			for(int i=1;i<=ny;i++)
			{
				if(!visy[i]&&slock[i]<d)
					d=slock[i];
			}
			for(int i=1;i<=nx;i++)
			{
				if(visx[i])
				lx[i]-=d;
			}
			for(int i=1;i<=ny;i++)
			{
				if(visy[i])
					ly[i]+=d;
				else
					slock[i]-=d;
			}
		}
	}
	int ans = 0;
	for(int i = 1;i <= ny; i++)
	ans += map[match[i]][i];
	printf("%d\n",-ans);
}
int main()
{
	while(scanf("%d%d",&N,&M),N||M)
	{
		getMap();
		KM();
	}
	return 0;
}