POJ 2485 Highways Kruskal+并查集最小生成树

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POJ 2485 Highways Kruskal+并查集最小生成树

2013年10月26日 ⁄ 综合 ⁄ 共 2588字 ⁄ 字号小中大 ⁄ 评论关闭

http://poj.org/problem?id=2485

由于本题是多CASE，全局变量忘记重置了，唉，WA了好多次！！！

Highways

Time Limit: 1000MS

Memory Limit: 65536K

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

/* Author : yan
 * Question : POJ 2485 Highways
 * Date && Time : Wednesday, February 16 2011 04:28 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
#define MAX 501
int map[MAX][MAX];
int father[MAX];
int rank[MAX];
typedef struct _route
{
	int x,y;
	int weight;
}Route;
Route route[MAX*MAX/2];
int route_cnt;
int ans;

void initial()
{
	ans=-1;
	route_cnt=0;
	unsigned int i;
	memset(rank,0,sizeof(rank));
	for(i=0;i<MAX;i++) father[i]=i;
}
int Find_Set(int x)
{
	if(x!=father[x]) father[x]=Find_Set(father[x]);
	return father[x];
}
int Union_Set(int x,int y,int w)
{
	int a=Find_Set(x);
	int b=Find_Set(y);
	if(a==b) return 0;
	if(rank[x]<rank[y])
	{
		father[a]=b;
	}
	else
	{
		father[b]=a;
		if(rank[x]==rank[y]) rank[x]++;
	}
	ans=ans>w?ans:w;
	return 1;
}
int cmp(const void *a,const void *b)
{
	return (*(Route *)a).weight - (*(Route *)b).weight;
}
int main()
{
	//freopen("input","r",stdin);
	unsigned int n,i,j,test;
	scanf("%d",&test);
	while(test--)
	{
		initial();
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++) scanf("%d",&map[i][j]);
		}
		for(i=0;i<n;i++)
		{
			for(j=i+1;j<n;j++)
			{
				route[route_cnt].x=i;
				route[route_cnt].y=j;
				route[route_cnt].weight=map[i][j];
				route_cnt++;
			}
		}
		qsort(route,route_cnt,sizeof(route[0]),cmp);

		for(i=0;i<route_cnt;i++)
		{
			Union_Set(route[i].x,route[i].y,route[i].weight);
			//printf("%d %d %d/n",route[i].x,route[i].y,route[i].weight);
		}
		printf("%d/n",ans);
	}
	return 0;
}

【上篇】kde汉化
【下篇】关于position中的relative

作者: murky

该日志由 murky 于11年前发表在综合分类下，最后更新于 2013年10月26日.
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