Highways

最新推荐文章于 2017-03-14 22:36:31 发布

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本文介绍了一个使用Prim算法解决的问题——如何在确保所有村庄通过公路相连的前提下，最小化最长公路的长度。具体介绍了输入输出格式、样例及算法实现。

Highways

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29349		Accepted: 13368

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

Sample Output

Hint

Huge input,scanf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU
和上一道题一样，直接用prim算法，注意用scanf输入

#include<stdio.h>
#define inf (1 << 20)
int dis[550][550];
void prim(int n)
{
    int ans = 0,minm,lowp[550],f;
    for(int i = 0;i < n;i++)
    {
        lowp[i] = dis[0][i];
    }
    for(int i = 0;i < n-1;i++)
    {
         minm = inf;
         for(int j = 0;j < n;j++)
         {
             if(lowp[j] && minm > lowp[j])
             {
                 minm = lowp[j];
                 f = j;
             }
         }
         if(ans < minm)
             ans = minm;
         lowp[f] = 0;
         for(int j = 0;j < n;j++)
         {
             if(lowp[j] > dis[f][j])
             {
                 lowp[j] = dis[f][j];
             }
         }
    }
    printf("%d\n",ans);
}
int main()
{
   int t,n;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%d",&n);
       for(int i = 0;i < n;i++)
       {
           for(int j = 0;j < n;j++)
           {
               scanf("%d",&dis[i][j]);
           }
       }
       prim(n);
   }
   return 0;
}