UVa 908

　　题目大意：有n个网站，由m条线路相连，每条线路都有一定的花费，找出连接所有线路的最小花费。

　　最小生成树问题(Minimal Spanning Tree, MST)，使用Kruskal算法解决。

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> ii;
#define MAXN 1000100

int p[MAXN];

int find(int x)
{
    return p[x] == x ? x : p[x] = find(p[x]);
}

int main()
{
#ifdef LOCAL
    freopen("in", "r", stdin);
#endif
    int n;
    bool first = true;
    while (scanf("%d", &n) != EOF)
    {
        int u, v, w;
        int old_cost = 0;
        for (int i = 0; i < n-1; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            old_cost += w;
        }
        int k;
        scanf("%d", &k);
        vector< pair<int, ii> > EdgeList;
        for (int i = 0; i < k; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            EdgeList.push_back(make_pair(w, make_pair(u, v)));
        }
        int m;
        scanf("%d", &m);
        for (int i = 0; i < m; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            EdgeList.push_back(make_pair(w, make_pair(u, v)));
        }
        sort(EdgeList.begin(), EdgeList.end());
        for (int i = 1; i <= n; i++)
            p[i] = i;
        int new_cost = 0;
        for (int i = 0; i < EdgeList.size(); i++)
        {
            w = EdgeList[i].first;
            u = EdgeList[i].second.first;
            v = EdgeList[i].second.second;
            int pu = find(u);
            int pv = find(v);
            if (pu != pv)
            {
                new_cost += w;
                p[pv] = pu;
            }
        }
        if (first)  first = false;
        else  printf("
");
        printf("%d
%d
", old_cost, new_cost);
    }
    return 0;
}