zoukankan      html  css  js  c++  java
  • 杭电1162Eddy's picture

    Eddy's picture

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 28   Accepted Submission(s) : 17

    Font: Times New Roman | Verdana | Georgia

    Font Size:

    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
    

    这道题也是非常easy的最短路径问题

    代码:

    #include<stdio.h>
    #include<string.h>
    #include<math.h>
    #define INF 1 << 30
    double a[101] , b[101] , map[101][101] ;
    double dis[101] ;
    int used[101] ;
    void prim(int n)
    {
     int c = 0 ;
     int i = 0 , j = 0 ;
     double sum = 0 ;
     dis[1] = 0 ;
     for(i = 1 ; i <= n ; i++)
     {
      double min = INF ;
      c = 0 ;
      for(j = 1 ; j <= n ; j++)
      {
             if(!used[j] && dis[j] < min)
       {
        min = dis[j] ;
        c =j ;
       }
      }
      used[c] = 1 ;
      for(j  = 1 ; j <= n ; j++ )
      {
       if(!used[j] && dis[j] > map[c][j])
        dis[j] = map[c][j] ;
      }
     }
     
     for(i = 1 ; i <= n ; i++)
      sum += dis[i] ;
     printf("%.2lf ",sum);
    }
    int main()
    {
     int n = 0 ;
     while(~scanf("%d",&n))
     {
      memset(a , 0 , sizeof( a ) ) ;
      memset(b , 0 , sizeof( b ) ) ;
      int i = 0 , j = 0 ;
      for(i = 1 ; i <= n ; i++)
      {
       for(j = 1 ; j <= n ; j++)
           map[i][j] = INF ;
       dis[i] = INF ;
       used[i] = 0 ;
      }
      for(i = 1 ; i <= n ; i++)
      {
                scanf("%lf%lf" , &a[i] , &b[i] );
      }
      double m = 0 , x = 0;
      for(i = 1 ; i <= n ; i++ )
      {
                for(j = 1 ; j <= n ; j++)
       {
        x = (a[j]-a[i])*(a[j]-a[i])+(b[j]-b[i])*(b[j]-b[i]) ;
        m = sqrt( x ) ;
        map[i][j] = map[j][i] = m ;
       }
      }
      prim( n );
     }
     return 0 ;
    }

  • 相关阅读:
    Python之struct模块浅谈
    看头发知健康
    ZeroMQ:云计算时代最好的通讯库
    粗盐热敷疗法经验汇总
    百度2011校招笔试算法题一
    new/delete 和malloc/free 的区别一般汇总
    Trie字典树
    百度2012校招笔试题之全排列与组合
    百度2011校招笔试算法题二
    执行程序的内存分布总结
  • 原文地址:https://www.cnblogs.com/mengfanrong/p/4223464.html
Copyright © 2011-2022 走看看