UVa 908 - Re-connecting Computer Sites

　　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 = ;
         for (int i = ; i < n-; i++)
         {
             scanf("%d%d%d", &u, &v, &w);
             old_cost += w;
         }
         int k;
         scanf("%d", &k);
         vector< pair<int, ii> > EdgeList;
         for (int i = ; 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 = ; 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 = ; i <= n; i++)
             p[i] = i;
         int new_cost = ;
         for (int i = ; 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("\n");
         printf("%d\n%d\n", old_cost, new_cost);
     }
     return ;
 }