poj 2195 Going Home

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
很显然的一个二分图，把人看作X，把房子看作Y，这样就构成了二分图，然后人和房子的距离就是人要多少步才能走到房子，这样就构成了一个带权的二分图，然后就是求二分图的最佳完美匹配，注意的是把距离弄成负的，因为求最大，然后跑一边KM算法就可以了，加油！！！这是我的第二道二分图最佳完美匹配题目

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f

using namespace std;

const double pi=acos(-1.0);
const double eps=1e-8;
typedef pair<int,int>pii;

const int maxn=100+10;

int W[maxn][maxn],n,c,m;//边的权值
int Lx[maxn],Ly[maxn];//顶标
int lleft[maxn];//右边第i个点的编号
bool S[maxn],T[maxn];//左右第i个点是否标记

struct node
{
    int x,y;
}people[maxn],house[maxn];

char str[maxn][maxn];

bool match(int i)//判断是否有增广路
{
    S[i]=true;
    for (int j=1;j<=n;j++) if (Lx[i]+Ly[j]==W[i][j] && !T[j])
    {
        T[j]=true;
        if (!lleft[j] || match(lleft[j]))
        {
            lleft[j]=i;
            return true;
        }
    }
    return false;
}

void update()//如果有增广路的话，更新
{
    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;
    }
}

void k_m()
{
    for (int i=1;i<=n;i++)
    {
        lleft[i]=Lx[i]=Ly[i]=0;
        for (int j=1;j<=n;j++)
        Lx[i]=max(Lx[i],W[i][j]);//初始可行顶标
    }
    for (int i=1;i<=n;i++)
    {
        for (; ;)
        {
            for (int j=1;j<=n;j++)
            S[j]=T[j]=0;//开始都没有标记过
            if (match(i)) break;
            else update();
        }
    }
}

int find_cost()
{
    int ans=0;
    for (int i=1;i<=n;i++)
    ans+=0-W[lleft[i]][i];
    return ans;
}

void init()
{
    int p=0,h=0;
    for (int i=0;i<c;i++)
    for (int j=0;j<m;j++)
    {
        if (str[i][j]=='H')
        {
            house[++h].x=i;
            house[h].y=j;
        }
        if (str[i][j]=='m')
        {
            people[++p].x=i;
            people[p].y=j;
        }
    }
    for (int i=1;i<=p;i++)
    for (int j=1;j<=h;j++)
    {
        W[i][j]=0-(abs(people[i].x-house[j].x)+abs(people[i].y-house[j].y));
    }
    n=p;
}

int main()
{
    //freopen("in.txt","r",stdin);
    while (scanf("%d%d",&c,&m)!=EOF && !(c==0 && m==0))
    {
        for (int i=0;i<c;i++) scanf("%s",str[i]);
        init();
        k_m();
        printf("%d
",find_cost());
    }
    //fclose(stdin);
    return 0;
}

我也发现，现在我现在写程序也有点规范了！！！！