zoukankan      html  css  js  c++  java
  • Highways POJ 2485【Prim】

    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



    #include<stdio.h>
    #include<string.h>
    #define INF 0x3f3f3f3f
    #define N 550
    int n;
    int i,j;
    int map[N][N];
    int a,b;
    int low[N];
    bool vis[N];
    int sum[N];
    void input()
    {
    	//memset(map,INF,sizeof(map));
    	for(i=1;i<=n;++i)
    	{
    		for(j=1;j<=n;++j)
    		{
    			scanf("%d",map[i]+j);
    		}
    	}
    }
    void prim()
    {
    	int pos=1;
    	for(i=1;i<=n;++i)//第一次给low赋值 
    	{
    		low[i]=map[pos][i];
    	}
    	vis[pos]=1; //增加最小生成树集合 
    	for(i=1;i<n;++i)//再找n-1个点 
    	{
    		int min=INF;
    		for(j=1;j<=n;++j)
    		{
    			if(!vis[j]&&min>low[j])
    			{
    				min=low[j];
    				pos=j;//把找到的点记录下
    			}
    		}
    		sum[i]=min;
    		vis[pos]=1; 
    		for(j=1;j<=n;++j)
    		{
    			if(!vis[j]&&low[j]>map[pos][j])
    			{
    				low[j]=map[pos][j];
    			}
    		} 
    	}
    }
    int main()
    {
    	int T;
    	scanf("%d",&T);
    	while(T--)
    	{
    		scanf("%d",&n);
    		memset(vis,0,sizeof(vis));
    		input();
    		prim();
    		int max=-INF;
    		for(i=1;i<n;++i)
    			if(max<sum[i])
    				max=sum[i];
    		printf("%d
    ",max);
    	}
    	return 0;
    }


  • 相关阅读:
    Oracle第三方ado.net数据提供程序
    HTML5实战 文摘 第二章 HTML5用于创建表单的输入小部件 数据绑定以及数据验证
    reactjs & antd & redux 使用心得
    HTML5实战 文摘 第一章 从文档到应用的转变
    css的高级选择器
    <!DOCTYPE html> 是什么,它的作用是什么
    Nginx调优
    java enum 枚举简单用法
    简谈java解析HTML(org.jsoup.nodes.Document)
    jquery EasyUI tree 加载远程数据 java实现
  • 原文地址:https://www.cnblogs.com/mfmdaoyou/p/7191154.html
Copyright © 2011-2022 走看看