Going Home
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5405 Accepted Submission(s): 2830
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
/*其实在求最大 最小的时候只要用一个模板就行了,把边的权值去相反数即可得到另外一个.求结果的时候再去相反数即可*/
/*最大最小有一些地方不同。。*/
#include<string.h>
#include<stdio.h>
#include<math.h>int fx,fy;
const int maxn = 301;
const int INF = 0xffffff;
struct node
{
int x,y;
}sx[maxn],sy[maxn];
int w[maxn][maxn];
int lx[maxn],ly[maxn]; //顶标int linky[maxn];
int visx[maxn],visy[maxn];
int slack[maxn];
bool find(int x)
{
visx[x] = true;
for(int y = 1; y <=fy; y++)
{
if(visy[y])
continue;
int t = lx[x] + ly[y] - w[x][y];
if(t==0)
{
visy[y] = true;
if(linky[y]==-1 || find(linky[y]))
{
linky[y] = x;
return true; //找到增广轨 }
}
else if(slack[y] > t)
slack[y] = t;
}
return false; //没有找到增广轨（说明顶点x没有对应的匹配，与完备匹配(相等子图的完备匹配)不符）}
int KM() //返回最优匹配的值{
int i,j;
memset(linky,-1,sizeof(linky));
memset(ly,0,sizeof(ly));
for(i = 1; i <=fx; i++)
{
lx[i] = -INF;
for(j = 1; j <=fy; j++)
if(w[i][j] > lx[i])
lx[i] = w[i][j];
}
for(int x = 1; x <=fx; x++)
{
for(i = 1; i <=fy; i++)
slack[i] = INF;
while(true)
{
memset(visx,0,sizeof(visx));
memset(visy,0,sizeof(visy));
if(find(x)) //找到增广轨，退出 break;
int d = INF;
for(i = 1; i <=fy; i++) //没找到，对l做调整(这会增加相等子图的边)，重新找 {
if(!visy[i] && d > slack[i])
d = slack[i];
}
for(i = 1; i <=fx; i++)
{
if(visx[i])
lx[i] -= d;
}
for(i = 1; i <=fy; i++)
{
if(visy[i])
ly[i] += d;
else
slack[i] -= d;
}
}
}
int result = 0;
for(i = 1; i <=fy; i++)
if(linky[i]>-1)
result += w[linky[i]][i];
return result;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m),n+m)
{
char cost[maxn];
int i,j;fx=0,fy=0;
for( i=0;i<n;i++)
{
scanf("%s",cost);
for(int j=0;j<m;j++)
if(cost[j]=='m')
sx[++fx].x=i,sx[fx].y=j;
else if(cost[j]=='H')
sy[++fy].x=i,sy[fy].y=j;
}
for(i=1;i<=fx;i++)
{
for(j=1;j<=fy;j++)
{
int xx=fabs(sx[i].y-sy[j].y)+fabs(sx[i].x-sy[j].x);
w[i][j]=-xx;
}
}
printf("%d
",-KM());
}
return 0;
}