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。

 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("\n");
65         printf("%d\n%d\n", old_cost, new_cost);
66     }
67     return 0;
68 }
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posted @ 2013-09-17 20:08  xiaobaibuhei  阅读(361)  评论(0)  编辑  收藏  举报