hdu 1533 Going Home（KM算法）

Going Home

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1415    Accepted Submission(s): 698

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
  
  

 

 

 

           题目大意：输入一幅图，图中有n个H和n个m分别代表有n个人和n个住所，人去住所需要花费金钱，移动一格需要一个费用，一格住所只能容纳一个人，问n个人都去到住所最低花费是多少？

求最低费用时，只要讲费用全部变成负数，然后按照原来那样求最大值，得出的结果再变成正数就是最后答案了，求最大值和最小值都是一样道理。

               赤裸裸的最大权匹配问题，先建好图，然后一个模板套上去就OK了。

代码如下：

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
using namespace std;

#define MAXN 100
#define INF 99999999

struct node
{
    int x, y;
}M[MAXN], H[MAXN];

char map[MAXN][MAXN];
int w[MAXN][MAXN];
int lx[MAXN], ly[MAXN];
bool sx[MAXN], sy[MAXN];
int pre[MAXN], slack[MAXN];
int n, m, mn, hn;

int dis(node a, node b)
{
    int dis = abs(a.x-b.x) + abs(a.y-b.y);
    return dis;
}

void bulid()
{
    int i, j;
    mn = hn = 0;
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < m; j++)
        {
            cin >> map[i][j];
            if(map[i][j] == 'm')
            {
                M[mn].x = i;
                M[mn].y = j;
                mn++;
            }
            if(map[i][j] == 'H')
            {
                H[hn].x = i;
                H[hn].y = j;
                hn++;
            }
        }
    }
    memset(w, 0, sizeof(w));
    for(i = 0; i < mn; i++)
        for(j = 0; j < hn; j++)
            w[i][j] = -dis(M[i], H[j]);
}

int dfs(int u)
{
    sx[u] = true;
    for(int i = 0; i < hn; i++)
    {
        if(sy[i]) continue;
        int tp = lx[u] + ly[i] - w[u][i];
        if(!tp)
        {
            sy[i] = true;
            if(pre[i] == -1 || dfs(pre[i]))
            {
                pre[i] = u;
                return 1;
            }
        }
        else if(slack[i] > tp) slack[i] = tp;
    }
    return 0;
}

int KM()
{
    int i, j, k, d, res = 0;
    memset(pre, -1, sizeof(pre));
    memset(ly, 0, sizeof(ly));
    for(i = 0; i < mn; i++)
    {
        lx[i] = -INF;
        for(j = 0; j < hn; j++)
            lx[i] = max(lx[i], w[i][j]);
    }
    for(k = 0; k < mn; k++)
    {
        for(i = 0; i < hn; i++)
            slack[i] = INF;
        while(1)
        {
            memset(sx, false, sizeof(sx));
            memset(sy, false, sizeof(sy));
            if(dfs(k)) break;
            d = INF;
            for(i = 0; i < hn; i++)
                if(!sy[i]) d = min(d, slack[i]);
            for(i = 0; i < mn; i++)
                if(sx[i]) lx[i] -= d;
            for(i = 0; i < hn; i++)
                if(sy[i]) ly[i] += d;
        }
    }
    res = 0;
    for(i = 0; i < hn; i++)
        res += w[pre[i]][i];
    return -res;
}

int main()
{
    int ans;
    while(scanf("%d %d", &n, &m) != EOF)
    {
        if(!n && !m) break;
        bulid();
        ans = KM();
        printf("%d\n", ans);
    }

    return 0;
}