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[关键字]：图论 二分图最佳匹配

[题目大意]：给出n*m的地图，由几个H（房子）和m（人），求每个人都走到一个房子的最少需要的总步数。

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[分析]：这道题是二分图最大匹配的应用。先把边权取负然后就是求一个边权最大的匹配。利用KM算法求解，具体：http://blog.sina.com.cn/s/blog_625c774e0100hhh7.html

[代码]：

View Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int MAXN=110;
const int INF=0x7fffffff;

int n,m,house,man;
int xh[MAXN],yh[MAXN],xm[MAXN],ym[MAXN];
int a[MAXN][MAXN],slack[MAXN],lx[MAXN],ly[MAXN],match[MAXN];
bool vx[MAXN],vy[MAXN];
char map[MAXN][MAXN];

bool Init()
{
    scanf("%d%d",&n,&m);
    if (!n && !m) return 0;
    for (int i=0;i<n;++i) scanf("%s",map[i]);
    house=man=0;
    for (int i=0;i<n;++i)
        for (int j=0;j<m;++j)
        {
            if (map[i][j]=='H') xh[++house]=i,yh[house]=j;
            if (map[i][j]=='m') xm[++man]=i,ym[man]=j;
        }
    for (int i=1;i<=man;++i)
        for (int j=1;j<=house;++j)
            a[i][j]=-(abs(xm[i]-xh[j])+abs(ym[i]-yh[j]));
    return 1;
}

bool DFS(int k)
{
    vx[k]=1;
    for (int i=1;i<=house;++i)
        if (!vy[i])
        {
            int temp=lx[k]+ly[i]-a[k][i];
            if (!temp)
            {
                vy[i]=1;
                if (!match[i] || DFS(match[i]))
                {
                    match[i]=k;
                    return 1;
                }
            }
            else 
                if (slack[i]>temp) slack[i]=temp;
        }
    return 0;
}

void Solve()
{
    memset(match,0,sizeof(match));
    memset(lx,200,sizeof(lx));
    for (int i=1;i<=man;++i)
        for (int j=1;j<=house;++j)
            if (lx[i]<a[i][j]) lx[i]=a[i][j];
    memset(ly,0,sizeof(ly));
    for (int i=1;i<=man;++i)
    {
        memset(slack,100,sizeof(slack));
        while (1)
        {
            memset(vx,0,sizeof(vx));
            memset(vy,0,sizeof(vy));
            if (DFS(i)) break;
            int Min=INF;
            for (int j=1;j<=house;++j)
                if (!vy[j] && Min>slack[j]) Min=slack[j];
            for (int j=1;j<=man;++j)
                if (vx[j]) lx[j]-=Min;
            for (int j=1;j<=house;++j)
                if (vy[j]) ly[j]+=Min; else slack[j]-=Min;
        }
    }
    int ans=0;
    for (int i=1;i<=house;++i)
        ans+=a[match[i]][i];
    printf("%d\n",-ans);
}

int main()
{
    while (Init()) Solve();
    return 0;
}