zoukankan      html  css  js  c++  java
  • POJ 2485:Highways(最小生成树&&prim)

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 21628   Accepted: 9970

    Description

    The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that it will be possible to drive between any pair of towns without leaving the highway system. 

    Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways. 

    The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.

    Input

    The first line of input is an integer T, which tells how many test cases followed. 
    The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between village i and village j. There is an empty line after each test case.

    Output

    For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.

    Sample Input

    1
    
    3
    0 990 692
    990 0 179
    692 179 0

    Sample Output

    692


    这题写的真烦。

    各种不知道的莫名其妙的

    
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<vector>
    #include<queue>
    #include<cmath>
    
    using namespace std;
    
    const int INF=1000000000;
    const int N=510;
    int map[N][N];
    int vis[N];
    int ans;
    int dis[N];
    
    void prim(int n)//prim求最小生成树
    {
        memset(vis, 0, sizeof(vis));
        for(int i=1; i<=n; i++)
        {
            dis[i] = INF;
        }
        dis[1] = 0;
        ans = 0;
        for(int i=1; i<=n; i++)
        {
            int temp = INF, k = 0;
            for(int j=1; j<=n; j++)
            {
                if(!vis[j] && dis[j]<temp)
                {
                    temp = dis[j];
                    k = j;
                }
            }
            vis[k] = true;
            if(ans<temp) ans = temp;
            for(int j=1; j<=n; j++)
            {
                if(!vis[j] && dis[j]>map[k][j])
                {
                    dis[j] = map[k][j];
                }
            }
        }
    }
    
    int main()
    {
        int t;
        scanf("%d", &t);
        while(t--)
        {
           int n;
           scanf("%d", &n);
           for(int i=1; i<=n; i++)
           {
              for(int j=1; j<=n; j++)
              {
                  scanf("%d", &map[i][j]);
              }
           }
           prim(n);
           printf("%d
    ", ans);
        }
        return 0;
    }
    

    错误。。







    版权声明:本文博客原创文章,博客,未经同意,不得转载。

  • 相关阅读:
    Win7下IE 8内存保护可能导致ActiveX无法安装的问题及其解决方法
    为Windows Server 2000开发和部署ASP.NET 3.5的应用程序
    MOSS 2010:如何为列表设计Infopath表单用于编辑和显示
    Windows Live is designed for you, but maybe not for your browser
    Meet the new WSS SharePoint Foundation 2010
    MOSS 2010: Visual Studio 2010开发体验(3)——调试代码
    MOSS 2010:Visual Studio 2010开发体验(10)——列表开发之内容类型
    MOSS 2010:谁动了我的“共享服务”
    MOSS 2010:通过SharePoint Designer定制列表项的条件格式
    如何访问嵌套母版页中的控件
  • 原文地址:https://www.cnblogs.com/hrhguanli/p/4753247.html
Copyright © 2011-2022 走看看