[ACM] HDU 1533 Going Home （二分图最小权匹配，KM算法）

Posted 2020-09-18 lxjshuju

Going Home

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2963    Accepted Submission(s): 1492

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

 

解题思路：

题意为n*m的方格中有p个房子和p个人，人每走一步花费1美元，要求每一个人都要找到一个房子，且每一个房子里仅仅能有一个人，问p个人都找到各自的房子。最小的花费是多少。

能够用KM算法求出最大的花费，要求最小花费仅仅要把邻接矩阵中的权值取反。然后用Km算法最后返回 最大值的相反数就是要求的最小花费。

代码：

#include <iostream>
#include <string.h>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=102;
const int inf=0x3f3f3f3f;
char mp[maxn][maxn];
int n,m;
int npeople;
int g[maxn][maxn];
int nx,ny;
int linked[maxn],lx[maxn],ly[maxn];
int slack[maxn];
bool visx[maxn],visy[maxn];

struct House//存放房子的坐标
{
    int x,y;
}house[maxn];

struct Man//存放人的坐标
{
    int x,y;
}man[maxn];

bool DFS(int x)//hungary
{
    visx[x]=true;
    for(int y=0;y<ny;y++)
    {
        if(visy[y])
            continue;
        int tmp=lx[x]+ly[y]-g[x][y];
        if(tmp==0)
        {
            visy[y]=true;
            if(linked[y]==-1||DFS(linked[y]))
            {
                linked[y]=x;
                return true;
            }
        }
        else if(slack[y]>tmp)
            slack[y]=tmp;
    }
    return false;
}

int KM()
{
    memset(linked,-1,sizeof(linked));
    memset(ly,0,sizeof(ly));
    for(int i=0;i<nx;i++)
    {
        lx[i]=-inf;
        for(int j=0;j<ny;j++)
            if(g[i][j]>lx[i])
            lx[i]=g[i][j];
    }
    for(int x=0;x<nx;x++)
    {
        for(int i=0;i<ny;i++)
            slack[i]=inf;
        while(true)
        {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if(DFS(x))
                break;
            int d=inf;
            for(int i=0;i<ny;i++)
                if(!visy[i]&&d>slack[i])
                d=slack[i];
            for(int i=0;i<nx;i++)
                if(visx[i])
                lx[i]-=d;
            for(int i=0;i<ny;i++)
            {
                if(visy[i])
                    ly[i]+=d;
                else
                    slack[i]-=d;
            }
        }
    }
    int ans=0;
    for(int i=0;i<ny;i++)
        if(linked[i]!=-1)
        ans+=g[linked[i]][i];
    return -ans;//返回最大的相反数。即要求的最小
}


int main()
{
    while(cin>>n>>m&&(n||m))
    {
        int h1=-1,m1=-1;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
        {
            cin>>mp[i][j];
            if(mp[i][j]==‘H‘)
                house[++h1].x=i,house[h1].y=j;
            else if(mp[i][j]==‘m‘)
                man[++m1].x=i,man[m1].y=j;
        }
        npeople=(++h1);
        memset(g,inf,sizeof(g));
        for(int i=0;i<npeople;i++)//计算邻接矩阵
            for(int j=0;j<npeople;j++)
            {
                int dist=abs(man[i].x-house[j].x)+abs(man[i].y-house[j].y);
                g[i][j]=-dist;//权值取反
            }
        nx=ny=npeople;
        cout<<KM()<<endl;
    }
    return 0;
}