POJ-2485 Highways 最小生成树

最小生成树的最大边长度

原创于 2021-07-28 18:40:19 发布 · 168 阅读

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该博客主要讨论如何解决图论问题中的一类经典题目，即在一个岛屿国家Flatopia中规划高速公路网络，使得所有城镇都能通过高速公路直接或间接到达，并且使最长的公路长度最小。博主介绍了Prim算法的应用，用于找到最小生成树并找出其中的最大边，从而给出最小可能的最长公路长度。此问题涉及到图的遍历、最短路径和最小生成树的概念，对于理解图算法和优化问题有一定的参考价值。

Highways

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 42418 Accepted: 18450

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
Hint

Huge input,scanf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

题意：求最小生成树里面最大的一条边

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int n,m;int ans;int x1,x2;
int a[510][510],book[510],dis[510];
int prim()
{
	int u;
	
	for(int i=1;i<=n;i++)
	{
		book[i]=0;
		dis[i]=a[1][i];
	}
	book[1]=1;
	dis[1]=0;
	
	for(int i=1;i<n;i++)
	{
		int minn=inf;
		
		for(int j=1;j<=n;j++)
		{
			if(dis[j]<minn&&book[j]==0)
			{
				minn=dis[j];
				u=j;
			}
		}
		book[u]=1;
		if(dis[u]>ans)ans=dis[u];
		for(int v=1;v<=n;v++)
		{
			if(book[v]==0&&dis[v]>a[u][v])
			dis[v]=a[u][v];
		
		}
	}
	return ans;
}
int main()
{
	int t;scanf("%d",&t);
	while(t--)
	{
		ans=0;
		scanf("%d",&n);
		memset(a,inf,sizeof(a));
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				scanf("%d",&a[i][j]);
			}
	    }
	    
		prim();
		
		printf("%d\n",ans);
	}
	return 0;
}