• Hdu 1533 Going Home


    Going Home

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 5405    Accepted Submission(s): 2830

    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

    /*其实在求最大 最小的时候只要用一个模板就行了,把边的权值去相反数即可得到另外一个.求结果的时候再去相反数即可*/
    /*最大最小有一些地方不同。。*/
    #include<string.h>
    #include<stdio.h>
    #include<math.h>int fx,fy;
    const int maxn = 301;
    const int INF = 0xffffff;
    struct node
    {
        int x,y;
    }sx[maxn],sy[maxn];
    int w[maxn][maxn];
    int lx[maxn],ly[maxn]; //顶标int linky[maxn];
    int visx[maxn],visy[maxn];
    int slack[maxn];
    
    bool find(int x)
    {
        visx[x] = true;
        for(int y = 1; y <=fy; y++)
        {
            if(visy[y])
                continue;
            int t = lx[x] + ly[y] - w[x][y];
            if(t==0)
            {
                visy[y] = true;
                if(linky[y]==-1 || find(linky[y]))
                {
                    linky[y] = x;
                    return true;        //找到增广轨            }
            }
            else if(slack[y] > t)
                slack[y] = t;
        }
        return false;                   //没有找到增广轨(说明顶点x没有对应的匹配,与完备匹配(相等子图的完备匹配)不符)}
    
    int KM()                //返回最优匹配的值{
        int i,j;
    
        memset(linky,-1,sizeof(linky));
        memset(ly,0,sizeof(ly));
        for(i = 1; i <=fx; i++)
        {
            lx[i] = -INF;
             for(j = 1; j <=fy; j++)
                if(w[i][j] > lx[i])
                    lx[i] = w[i][j];
        }
    
        for(int x = 1; x <=fx; x++)
        {
            for(i = 1; i <=fy; i++)
                slack[i] = INF;
            while(true)
            {
                memset(visx,0,sizeof(visx));
                memset(visy,0,sizeof(visy));
                if(find(x))                     //找到增广轨,退出                break;
                int d = INF;
                for(i = 1; i <=fy; i++)          //没找到,对l做调整(这会增加相等子图的边),重新找            {
                    if(!visy[i] && d > slack[i])
                        d = slack[i];
                }
                for(i = 1; i <=fx; i++)
                {
                    if(visx[i])
                        lx[i] -= d;
                }
                for(i = 1; i <=fy; i++)
                {
                    if(visy[i])
                         ly[i] += d;
                    else
                        slack[i] -= d;
                }
            }
        }
        int result = 0;
        for(i = 1; i <=fy; i++)
        if(linky[i]>-1)
            result += w[linky[i]][i];
        return result;
    }
    
    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m),n+m)
        {
            char cost[maxn];
            int i,j;fx=0,fy=0;
            for( i=0;i<n;i++)
            {
                 scanf("%s",cost);
                 for(int j=0;j<m;j++)
                 if(cost[j]=='m')
                 sx[++fx].x=i,sx[fx].y=j;
                 else if(cost[j]=='H')
                 sy[++fy].x=i,sy[fy].y=j;
            }
            for(i=1;i<=fx;i++)
            {
               for(j=1;j<=fy;j++)
               {
                  int xx=fabs(sx[i].y-sy[j].y)+fabs(sx[i].x-sy[j].x);
                  w[i][j]=-xx;
               }
     
            }
    
            printf("%d
    ",-KM());
        }
        return 0;
    }
    

      

  • 相关阅读:
    MySql错误解决方案汇总
    不适合做管理的人zz
    linux 自动执行 crontab学习笔记
    Google Megastore分布式存储技术全揭秘zz
    【算法】n个人围成一圈报数,报到3的退出,下面接着从1开始报,问最后剩下的是谁?
    大数据下的数据分析平台架构zz
    ETL的可扩展性和可维护性zz
    【算法】各种排序算法测试代码
    谈爱情故事,谈观察者模式
    解读设计模式单例模式(Singleton Pattern)
  • 原文地址:https://www.cnblogs.com/zhangliu/p/7058017.html
Copyright © 2020-2023  润新知