POJ 1861 Networks 最小生成树 Kruskal+并查集

这题就是最小生成树，不过样例输出看了好久，一看DISCUSS才知道，这题是SPJ，答案不唯一，不必管样例

下面Kruskal+并查集

一次AC！！

Network

Time Limit:1000MS		Memory Limit:30000K
Total Submissions:7721		Accepted:2835		Special Judge

Description

Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.

Input

The first line of the input contains two integer numbers: N - the number of hubs in the network (2 <= N <= 1000) and M - the number of possible hub connections (1 <= M <= 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 10⁶. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs.

Output

Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.

Sample Input

4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1

Sample Output

1
4
1 2
1 3
2 3
3 4

/* Author yan
 * POJ 1861
 * Network
*/
#include<stdio.h>
#define MAX 1000
typedef struct _route
{
	int x,y;
	int weight;
};
struct _route route[15000];
int cache[2][15000];//暂存节点
int count;//统计暂存节点
int n;
int father[MAX+1];
int rank[MAX+1];
int max_used;
int sum;

void initial()
{
	int _i;
	memset(rank,0,sizeof(rank));
	for(_i=0;_i<=n;_i++) father[_i]=_i;
	max_used=0;
	sum=0;
	count=0;
}
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,b;
	a=Find_Set(x);
	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]++;
	}
	max_used=w;
	cache[0][count]=x,cache[1][count]=y;
	count++;
	return 1;
}
int cmp(const void *a,const void *b)
{
	return (*(struct _route *)a).weight - (*(struct _route *)b).weight;
}
int main()
{
	int x,y,m;
	int i,j;
	//freopen("input","r",stdin);
	scanf("%d %d",&n,&m);
	initial();
	for(i=0;i<m;i++)
		scanf("%d %d %d",&route[i].x,&route[i].y,&route[i].weight);
	qsort(route,m,sizeof(route[0]),cmp);
	for(i=0;i<m;i++)
	{
/*		x=Find_Set(route[i].x);
		y=Find_Set(route[i].y);
		if(x!=y) Union_Set(route[i].x,route[i].y,route[i].weight);
*/
		Union_Set(route[i].x,route[i].y,route[i].weight);
	}
	printf("%d/n%d/n",max_used,count);
	for(i=0;i<count;i++)
		printf("%d %d/n",cache[0][i],cache[1][i]);
	return 0;
}