zoukankan      html  css  js  c++  java
  • Agri-Net prime算法

    Problem Description
    Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course. 
    Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms. 
    Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm. 
    The distance between any two farms will not exceed 100,000. 
     

    Input
    The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.
     

    Output
    For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.
     

    Sample Input
    4 0 4 9 21 4 0 8 17 9 8 0 16 21 17 16 0
     

    Sample Output
    28
    **************************************************************************************************************************
    最小生成树
    **************************************************************************************************************************
     1 #include<iostream>
     2 #include<string>
     3 #include<cmath>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<queue>
     7 using namespace std;
     8 int map[2010][2010];
     9 int vis[2010],n;
    10 int prime()
    11 {
    12     int lowcost[2010],lowset[2010];
    13     int it,jt,kt,sum=0;
    14     memset(vis,0,sizeof(vis));
    15     for(it=1;it<=n;it++)//初始化
    16     {
    17         lowcost[it]=map[1][it];
    18         //cout<<"lowc::"<<lowcost[it]<<endl;
    19     }
    20     vis[1]=1;
    21     for(it=1;it<n;it++)
    22     {
    23         jt=1;
    24         while(vis[jt])jt++;
    25         for(kt=1;kt<=n;kt++)//找到最短边
    26         {
    27             if(!vis[kt]&&lowcost[kt]<lowcost[jt])jt=kt;
    28         }
    29         //cout<<"j:: "<<jt<<endl;
    30         sum+=lowcost[jt];
    31         //cout<<"lowcost[jt]:: "<<lowcost[jt]<<endl;
    32         //cout<<"sum: "<<sum<<endl;
    33         vis[jt]=1;
    34         for(kt=1;kt<=n;kt++)//更新最短边
    35         {
    36             if(lowcost[kt]>map[jt][kt])
    37             {
    38                 lowcost[kt]=map[jt][kt];
    39                 //lowset[k]=j;
    40                 //cout<<"lowset: "<<lowset[kt]<<endl;
    41             }
    42         }
    43 
    44     }
    45     return sum;
    46 }
    47 int main()
    48 {
    49     int i,j,k;
    50     while(scanf("%d",&n)!=EOF)
    51     {
    52         for(i=1;i<=n;i++)
    53          for(j=1;j<=n;j++)
    54          {
    55             scanf("%d",&map[i][j]);
    56             //cout<<"map:: "<<map[i][j]<<endl;
    57          }
    58          printf("%d
    ",prime());
    59     }
    60     return 0;
    61 }
    View Code

  • 相关阅读:
    Java-使用IO流对大文件进行分割和分割后的合并
    Java-单向链表算法
    Java-二分查找算法
    Java-二叉树算法
    Java-对象比较器
    Android中Activity的四种开发模式
    Struts2工作原理
    C++实现单例模式
    数组中有一个数字出现的次数超过数组的一半,请找出这个数字
    c++ enum用法【转】
  • 原文地址:https://www.cnblogs.com/sdau--codeants/p/3379003.html
Copyright © 2011-2022 走看看