zoukankan      html  css  js  c++  java
  • HDOJ 1162

    Eddy's picture

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5797    Accepted Submission(s): 2913

    Problem Description
    Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you. Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?
     
    Input
    The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.
    Input contains multiple test cases. Process to the end of file.
     
    Output
    Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.
     
    Sample Input
    3 1.0 1.0 2.0 2.0 2.0 4.0
     
    Sample Output
    3.41
     
    核心算法:prim水题;
    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    double g[101][101];
    double ans;
    double min=0xfffff;
    void prim(int n)
    {
        double lowclose[101];
        int closet[101],used[101],i,j,k;
        memset(used,0,sizeof(used));
        for(i=1;i<=n;i++)
            lowclose[i]=g[i][1],
            closet[i]=1;
        used[1]=1;
        for(i=1;i<n;i++)
        {
            j=1;
            min=0xfffff;
            for(k=2;k<=n;k++)
            {
                if(lowclose[k]<min&&!used[k])
                    min=lowclose[k],
                    j=k;
            }
            used[j]=1;
            ans+=g[j][closet[j]];
            for(k=2;k<=n;k++)
            {
                if(g[k][j]<lowclose[k]&&!used[k])
                    lowclose[k]=g[k][j],
                    closet[k]=j;
            }
        }
    
    }
    double func(double x1,double y1,double x2,double y2)
    {
        return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    }
    int main()
    {
        int i,j,n;
        double t[101][2];
        while(scanf("%d",&n)!=EOF)
        {
            ans=0.0;
            for(i=1;i<=n;i++)
                g[i][i]=0xfffff,
                scanf("%lf %lf",&t[i][0],&t[i][1]);
            for(i=1;i<=n;i++)
            {
                for(j=1+i;j<=n;j++)
                {
                     g[j][i]=g[i][j]=func(t[i][0],t[i][1],t[j][0],t[j][1]);
                }
            }
            prim(n);
            printf("%.2lf
    ",ans);
        }
        return 0;
    }
    View Code
  • 相关阅读:
    IIS应用程序池标识(程序池账户)ApplicationPoolIdentify
    笔记二 sql 基础知识
    笔记三 索引讲解
    SqlParameter使用参数错误,提示请求超时或则查询速度过慢
    SVN 常见操作
    Excel 基本操作
    sql server row_number分页排序
    本地部署IIS
    sql中去掉字段的所有空格
    sql取逗号前后数据与批量修改某一字段某一值
  • 原文地址:https://www.cnblogs.com/zeze/p/hdoj1162.html
Copyright © 2011-2022 走看看