题目：有一些电脑，电脑间由一些线路连接。现在给你电脑间的连接状态，以及原来选取的网络线路。

            在此基础上加入了K条新线路，要从里面选取新的网络，使得电脑都连通并且线路长度和最小。

分析：图论、最小生成树。

说明：每组数据间有空行。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;

typedef struct d_node
{
	int point1;
	int point2;
	int weight;
}enode;
enode edge[1000010];

//union_set
int sets[1000010];
int rank[1000010];

void set_inital( int a, int b )
{
	for ( int i = a ; i <= b ; ++ i ) {
		rank[i] = 0;
		sets[i] = i;
	}
}

int  set_find( int a )
{
	if ( a != sets[a] )
		sets[a] = set_find( sets[a] );
	return sets[a];
}

void set_union( int a, int b )
{
	if ( rank[a] < rank[b] )
		sets[a] = b;
	else {
		if ( rank[a] == rank[b] )
			rank[a] ++;
		sets[b] = a;
	}
}
//end_union_set

int cmp_e( enode a, enode b )
{
	return a.weight < b.weight;
}

int kruskal( int n, int m )
{
	sort( edge, edge+m, cmp_e );
	
	set_inital( 0, n );
	int sum = 0;
	for ( int i = 0 ; i < m ; ++ i ) {
		int A = set_find( edge[i].point1 );
		int B = set_find( edge[i].point2 );
		if ( A != B ) { 
			set_union( A, B );
			sum += edge[i].weight;
		}
	}
	return sum;
}

int main()
{
	int N,K,M,u,v,w,c = 0;
	while ( ~scanf("%d",&N) ) {
		int old_cost = 0;
		for ( int i = 1 ; i < N ; ++ i ) {
			scanf("%d%d%d",&u,&v,&w); 
			old_cost += w;
		}
		
		int e_count = 0;
		scanf("%d",&K);
		for ( int i = 0 ; i < K ; ++ i ) {
			scanf("%d%d%d",&u,&v,&w);
			edge[e_count].point1 = u;
			edge[e_count].point2 = v; 
			edge[e_count].weight = w;
			e_count ++;
		}
		scanf("%d",&M);
		for ( int i = 0 ; i < M ; ++ i ) {
			scanf("%d%d%d",&u,&v,&w);
			edge[e_count].point1 = u;
			edge[e_count].point2 = v; 
			edge[e_count].weight = w;
			e_count ++;
		}
		
		if ( c ++ ) printf("\n");
		printf("%d\n%d\n",old_cost,kruskal( N, e_count ));
	}
	
	return 0;
}