Highways （最小生成树）

最新推荐文章于 2019-08-16 22:27:40 发布

While.True

最新推荐文章于 2019-08-16 22:27:40 发布

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分类专栏：最小生成树

本文链接：https://blog.youkuaiyun.com/qq_40912854/article/details/81750212

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本文介绍了一个使用最小生成树算法解决Flatopia国家交通规划问题的案例。Flatopia是一个地形平坦的岛国，面临缺乏公共高速公路的问题，导致交通不便。政府计划通过构建一些高速公路来改善这一状况，确保任何两个城镇间都能通过高速公路系统相连。文章详细阐述了如何通过读取村庄之间的距离，运用算法找出连接所有村庄的最小总长度的高速公路方案，同时确保最长的单条公路长度最小。

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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.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

struct node
{
	int x,y,len;
}road[500*500+5];

bool cmp(node a,node b)
{
	return a.len <b.len;
}

int a[505];

int find(int x)
{
	if (a[x]==x)
		return x;
	else
		return a[x] = find(a[x]);
}

bool unite(int x,int y)
{
	x = find(x);
	y = find(y);
	if (x!=y)
	{
		if (x<y)
			a[y] = x;
		else
			a[x] = y;
		return 1;
	}
	else
		return 0;
}

int main()
{
	int t;
	scanf("%d",&t);
	while (t--)
	{
		int n,i,j;
		scanf("%d",&n);
		for (i=1;i<=n;i++)
			a[i] = i;
		int k = 0;
		for (i=1;i<=n;i++)
			for (j=1;j<=n;j++)
			{
				road[k].x = i;
				road[k].y = j;
				scanf("%d",&road[k].len);
				k++;
			}
		sort(road,road+k,cmp);
		int sum,temp = 0;
		for (i=0;i<k;i++)
		{
			if (unite(road[i].x,road[i].y))
			{
				temp++;
				if (temp==n-1)
					sum = road[i].len;//如果是最后一条入边，那么这条边一定是权值最大边
			}
				
		}
		printf("%d\n",sum);
	}
	return 0;
}