HDU 1533 Going Home【km應用】

HDU 1533 Going Home【km應用】http://acm.hdu.edu.cn/showproblem.php?pid=1533

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

 

題目大意：

給你一個N行M列的矩陣，其中“.”代表空地，“H”代表房子，“m”代表人，其中有n個房子和n個人。現在要求每個人進入一間房子，且人走一步需要支付1美元。

求最小需要花費多少美元才能讓所有人都進入到房子中（每個人只能進入一間房子，每個房子只能容納一個人）。

解題思路：

這道題其實就是二分圖最優匹配的變形而已。

因爲要求的其實是最小權值之和。而KM算法求的是最大權值之和。

爲此，我們可以有兩種方法來解決：

1.在建圖的時候，將每條邊的權值變爲負數。然後lx[i]初始化爲-INT_MAX，結果輸出-ans，就可以得到最小權值。

2.在建圖的時候，用一個相對的大數減去每個權值，比如現在邊爲6,8,10.我用100來減，得到的邊爲94,92,90.最後輸出時，用100*n-ans也可以得到最小的權值。

方法1代碼如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <cmath>

using namespace std;

const int N = 310;
char mmap[N][N];
int map[N][N],slack[N],match[N],lx[N],ly[N];
bool visx[N],visy[N];
int n;

bool dfs(int x)
{
    visx[x] = true;
    for(int i=0; i<n; i++)
    {
        if(visy[i]) continue;
        int t = lx[x] + ly[i] - map[x][i];
        if(!t)
        {
            visy[i] = true;
            if(match[i] == -1 || dfs(match[i])){
                match[i] = x;
                return true;
            }
        }
        else
            slack[i] = min(slack[i], t);
    }
    return false;
}

void km()
{
    int temp;
    memset(lx,0,sizeof(lx));
    memset(ly,0,sizeof(ly));
    memset(match,-1,sizeof(match));
    for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
            lx[i] = max(lx[i],map[i][j]);
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++) slack[j] = INT_MAX;
        while(1)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(dfs(i)) break;
            else{
                temp = INT_MAX;
                for(int j=0; j<n; j++)
                {
                    if(!visy[j] && temp>slack[j])
                        temp = slack[j];
                }
                for(int j=0; j<n; j++)
                {
                    if(visx[j]) lx[j] -= temp;
                    if(visy[j]) ly[j] += temp;
                    else slack[j] -= temp;
                }
            }
        }
    }
}

int main()
{
    int row,col,ans,numi,numj;
    while(scanf("%d %d",&row,&col)!=EOF)
    {
        if(row == 0 && col == 0) break;
        n = ans = numi = numj = 0;
        for(int i=0; i<row; i++){
            scanf("%s",mmap[i]);
            for(int j=0; j<col; j++)
                if(mmap[i][j] == 'm') n++;
        }

        for(int i=0; i<row; i++){
            for(int j=0; j<col; j++){
                if(mmap[i][j] == 'm')
                {
                    for(int k=0; k<row; k++){
                        for(int l=0; l<col; l++){
                            if(mmap[k][l] == 'H')
                                map[numi][numj++] =  - (abs(k - i) + abs(l - j));//變爲負權邊
                        }
                    }
                    numi++;
                    numj = 0;
                }
            }
        }
        km();
        for(int i=0; i<n; i++)
            ans += map[ match[i] ][i];
        printf("%d\n", - ans);//結果取反
    }
    return 0;
}

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