zoukankan      html  css  js  c++  java
  • POJ 2195 Going Home

    Going Home
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 15130   Accepted: 7740

    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

       考察点: 最小费用最大流
    #include <stdio.h>
    #include <string.h>
    #include <math.h>
    struct num
    {
        int x,y;
    }h[200],m[200];
    int cost[300][300],cap[300][300];
    int queue[1000000],status[300],dis[300],pre[300];
    int INF=0x7fffffff,sta,end;
    int main()
    {
        int EK();
        int i,j,n,u,s,t,top1,top2,x1,y1,x2,y2;
        char c;
        while(scanf("%d %d%*c",&n,&u)!=EOF)
        {
            if(n==0&&u==0)
            {
                break;
            }
            top1=top2=0;
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=u;j++)
                {
                    scanf("%c",&c);
                    if(c=='H')
                    {
                        h[top1].x=i;
                        h[top1++].y= j;
                    }else if(c=='m')
                    {
                        m[top2].x= i;
                        m[top2++].y= j;
                    }
                }
                getchar();
            }
            if(top1==0)
            {
                printf("0\n");
                continue;
            }
            memset(cap,0,sizeof(cap));
            memset(cost,0,sizeof(cost));
            n=top1;
            for(i=1;i<=n;i++)
            {
                cap[0][i]=1;
            }
            for(i=1;i<=n;i++)
            {
                x1=m[i-1].x; y1=m[i-1].y;
                for(j=1;j<=n;j++)
                {
                    x2=h[j-1].x; y2=h[j-1].y;
                    x2= x2- x1;
                    if(x2<0)
                    {
                        x2=-x2;
                    }
                    y2=y2-y1;
                    if(y2<0)
                    {
                        y2=-y2;
                    }
                    cap[i][n+j]=1;
                    cost[i][n+j]=(x2+y2);
                    cost[n+j][i]=-(x2+y2);
                }
            }
            for(i=1;i<=n;i++)
            {
                cap[n+i][n+n+1]=1;
            }
            s=0;
            sta=0; end=2*n+1;
            s=EK();
            printf("%d\n",s);
        }
        return 0;
    }
    int bfs()
    {
        int i,j,base,top,x;
        base=top=0;
        for(i=0;i<=end;i++)
        {
            dis[i]=INF;
        }
        memset(status,0,sizeof(status));
        queue[top++]=0;
        status[0]=1;
        dis[0]=0;
        while(base<top)
        {
            x=queue[base++];
            status[x]=0;
            for(i=1;i<=end;i++)
            {
                if(x!=i&&cap[x][i])
                {
                    if(dis[i]>(dis[x]+cost[x][i]))
                    {
                        dis[i]=dis[x]+cost[x][i];
                        pre[i]=x;
                        if(!status[i])
                        {
                            status[i]=1;
                            queue[top++]=i;
                        }
                    }
                }
            }
        }
        if(dis[end]==INF)
        {
            return 0;
        }else
        {
            return 1;
        }
    }
    int EK()
    {
        int i,j,s=0,k,min=INF;
        while(1)
        {
            k=bfs();
            if(k==0)
            {
                break;
            }
            for(i=end; i!=0; i=pre[i])
            {
                j=pre[i];
                if(min>cap[j][i])
                {
                    min=cap[j][i];
                }
            }
            for(i=end; i!=0; i=pre[i])
            {
                j=pre[i];
                s+=(cost[j][i]*min);
                cap[j][i]-=min;
                cap[i][j]+=min;
            }
        }
        return s;
    }
    

  • 相关阅读:
    微信 JS SDK 的 chooseImage 接口在部分安卓机上容易造成页面刷新
    规约模式Specification Pattern
    ASP.NET Core 1.0基础之日志
    C# 7 新特性-2
    C# 7 新特性-1
    ASP.NET Core 1.0基础之诊断
    ASP.NET Core 1.0基础之依赖注入
    ASP.NET Core 1.0 基础之配置
    ASP.NET Core 1.0基础之静态文件处理
    FreeSql生产环境自动升级数据库解决方案
  • 原文地址:https://www.cnblogs.com/javawebsoa/p/3029476.html
Copyright © 2011-2022 走看看