POJ 2195 Going Home(二分图最优匹配)

题意：给定一个N*M的地图，地图上有若干个man和house，且man与house的数量一致。man每移动一格需花费$1（即单位费用=单位距离），一间house只能入住一个man。现在要求所有的man都入住house，求最小费用。

思路：之前用的网络流做的，现在用KM算法做。

           每个人对应二分图左边的点,每个房间对应二分图右边的点.每个人与每个房间都有一条带权值的边,表示这个人到该房间的花费.为了使得每个人都得到一个房间,且他们行走的开销最小.那么本题求得就是该二分图的一个完备匹配且该匹配边的权值和最小了.但是KM算法只能求二分图的最优匹配(即匹配边权值和最大的匹配),我们现在要求权值和最小的匹配怎么办?我们只需要把所有边的权值取负即可。最后结果再取负即可

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100+5;

struct Max_Match
{
    int W[maxn][maxn],n;
    int Lx[maxn],Ly[maxn];
    bool S[maxn],T[maxn];
    int left[maxn];

    bool match(int i)
    {
        S[i]=true;
        for(int j=1;j<=n;j++)if(Lx[i]+Ly[j]==W[i][j] && !T[j])
        {
            T[j]=true;
            if(left[j]==-1 || match(left[j]))
            {
                left[j]=i;
                return true;
            }
        }
        return false;
    }

    void update()
    {
        int a=1<<30;
        for(int i=1;i<=n;i++)if(S[i])
        for(int j=1;j<=n;j++)if(!T[j])
        {
            a = min(a,Lx[i]+Ly[j]-W[i][j]);
        }
        for(int i=1;i<=n;i++)
        {
            if(S[i]) Lx[i]-=a;
            if(T[i]) Ly[i]+=a;
        }
    }

    int solve(int n)
    {
        this->n=n;
        memset(left,-1,sizeof(left));
        for(int i=1;i<=n;i++)
        {
            Lx[i]=Ly[i]=0;
            for(int j=1;j<=n;j++)
                Lx[i]=max(Lx[i], W[i][j]);
        }

        for(int i=1;i<=n;i++)
        {
            while(true)
            {
                for(int j=1;j<=n;j++) S[j]=T[j]=false;
                if(match(i)) break;
                else update();
            }
        }

        int ans=0;
        for(int i=1;i<=n;i++) ans+= W[left[i]][i];
        return ans;
    }
}KM;

struct Node
{
    int x,y;
    Node(){}
    Node(int x,int y):x(x),y(y){}
}node1[maxn],node2[maxn];

int main()
{
    int R,C;
    while(scanf("%d%d",&R,&C)==2 && R)
    {
        int num1=0,num2=0;
        for(int i=1;i<=R;i++)
        for(int j=1;j<=C;j++)
        {
            char c;
            scanf(" %c",&c);
            if(c=='H') node2[++num2]=Node(i,j);
            else if(c=='m') node1[++num1]=Node(i,j);
        }

        for(int i=1;i<=num1;i++)
        for(int j=1;j<=num2;j++)
            KM.W[i][j]=-( abs(node1[i].x-node2[j].x)+abs(node1[i].y-node2[j].y) );//取负值
        int ans = KM.solve(num1);
        printf("%d\n",-ans);//取负值
    }
    return 0;
}

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