hdu1162 Eddy's picture 基础最小生成树 

 1     #include <cstdio>
 2     #include <cmath>
 3     #include <cstring>
 4     #include <algorithm>
 5     #include <string>
 6     using namespace std;
 7 
 8     int n, m;
 9     int f[510];
10 
11     struct node
12     {
13         int u, v;
14         double w;
15     }map[10010];
16 
17     int getf(int a)
18     {
19         if (f[a] == a)
20         {
21             return a;
22         }
23 
24         f[a] = getf(f[a]);
25         return f[a];
26     }
27 
28     void merge(int a, int b)
29     {
30         int x = getf(a);
31         int y = getf(b);
32 
33         if (x != y)
34             f[y] = x;
35     }
36 
37 
38     void init()
39     {
40         for (int i = 1; i <= n; i++)
41             f[i] = i;
42     }
43 
44     bool cmp(node a, node b)
45     {
46         return a.w < b.w;
47     }
48 
49     double kura()
50     {
51         double res = 0;
52         sort(map, map + m, cmp);
53 
54         for (int i = 0; i < m; i++)
55         {
56             if (getf(map[i].u) != getf(map[i].v))
57             {
58                 merge(map[i].u, map[i].v);
59                 res += map[i].w;
60             }
61         }
62         return res;
63     }
64 
65     double a[102][2];
66 
67 
68     int main()
69     {
70         while (scanf("%d", &n) != EOF)
71         {
72             m = n*(n - 1);
73             for (int i = 1; i <= n; i++)
74                 scanf("%lf %lf", &a[i][0], &a[i][1]);
75 
76             int k = 0;
77             for (int i = 1; i <= n; i++)
78             {
79                 for (int j = i + 1; j <= n; j++)
80                 {
81                     map[k].u = i;
82                     map[k].v = j;
83                     map[k].w = sqrt(abs(pow(a[j][0] - a[i][0], 2) + pow(a[j][1] - a[i][1], 2)));
84                     k++;
85                 }
86             }
87             init();
88             printf("%.2lf\n", kura());
89         }
90 
91         return 0;
92     }

 

posted @ 2017-08-06 18:32  ouyang_wsgwz  阅读(148)  评论(0)    收藏  举报