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。
1 #include <cstdio> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5 typedef pair<int, int> ii; 6 #define MAXN 1000100 7 8 int p[MAXN]; 9 10 int find(int x) 11 { 12 return p[x] == x ? x : p[x] = find(p[x]); 13 } 14 15 int main() 16 { 17 #ifdef LOCAL 18 freopen("in", "r", stdin); 19 #endif 20 int n; 21 bool first = true; 22 while (scanf("%d", &n) != EOF) 23 { 24 int u, v, w; 25 int old_cost = 0; 26 for (int i = 0; i < n-1; i++) 27 { 28 scanf("%d%d%d", &u, &v, &w); 29 old_cost += w; 30 } 31 int k; 32 scanf("%d", &k); 33 vector< pair<int, ii> > EdgeList; 34 for (int i = 0; i < k; i++) 35 { 36 scanf("%d%d%d", &u, &v, &w); 37 EdgeList.push_back(make_pair(w, make_pair(u, v))); 38 } 39 int m; 40 scanf("%d", &m); 41 for (int i = 0; i < m; i++) 42 { 43 scanf("%d%d%d", &u, &v, &w); 44 EdgeList.push_back(make_pair(w, make_pair(u, v))); 45 } 46 sort(EdgeList.begin(), EdgeList.end()); 47 for (int i = 1; i <= n; i++) 48 p[i] = i; 49 int new_cost = 0; 50 for (int i = 0; i < EdgeList.size(); i++) 51 { 52 w = EdgeList[i].first; 53 u = EdgeList[i].second.first; 54 v = EdgeList[i].second.second; 55 int pu = find(u); 56 int pv = find(v); 57 if (pu != pv) 58 { 59 new_cost += w; 60 p[pv] = pu; 61 } 62 } 63 if (first) first = false; 64 else printf(" "); 65 printf("%d %d ", old_cost, new_cost); 66 } 67 return 0; 68 }