hdu 1233 还是畅通工程 (prim， kruskal)

还是畅通工程
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58241 Accepted Submission(s): 26468

Problem Description
某省调查乡村交通状况，得到的统计表中列出了任意两村庄间的距离。省*“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通（但不一定有直接的公路相连，只要能间接通过公路可达即可），并要求铺设的公路总长度为最小。请计算最小的公路总长度。

Input
测试输入包含若干测试用例。每个测试用例的第1行给出村庄数目N ( < 100 )；随后的N(N-1)/2行对应村庄间的距离，每行给出一对正整数，分别是两个村庄的编号，以及此两村庄间的距离。为简单起见，村庄从1到N编号。
当N为0时，输入结束，该用例不被处理。

Output
对每个测试用例，在1行里输出最小的公路总长度。

Sample Input
3
1 2 1
1 3 2
2 3 4
4
1 2 1
1 3 4
1 4 1
2 3 3
2 4 2
3 4 5
0

Sample Output
3
5

Hint
Hint

Huge input, scanf is recommended.

C/C++(prim)：

 1 #include <cstdio>
 2 #include <climits>
 3 using namespace std;
 4 
 5 int n, my_map[110][110];
 6 
 7 int my_prim()
 8 {
 9     int my_pos = 1, my_book[110] = {0, 1}, my_dis[110] = {0, INT_MAX}, my_ans = 0;
10     for (int i = 2; i <= n; ++ i)
11         my_dis[i] = my_map[my_pos][i];
12     for (int i = 1; i < n; ++ i)
13     {
14         int my_temp = INT_MAX;
15         for (int j = 1; j <= n; ++ j)
16         {
17             if (!my_book[j] && my_dis[j] < my_temp)
18             {
19                 my_temp = my_dis[j];
20                 my_pos = j;
21             }
22         }
23         my_ans += my_temp;
24         my_book[my_pos] = 1;
25         for (int j = 1; j <= n; ++ j)
26             if (my_dis[j] > my_map[j][my_pos])
27                 my_dis[j] = my_map[j][my_pos];
28     }
29     return my_ans;
30 }
31 
32 int main()
33 {
34     while (scanf("%d", &n), n)
35     {
36         int my_line = n * (n-1) / 2;
37         for (int i = 0; i < my_line; ++ i)
38         {
39             int a, b, a_b_dis;
40             scanf("%d%d%d", &a, &b, &a_b_dis);
41             my_map[a][b] = my_map[b][a] = a_b_dis;
42         }
43         printf("%d
", my_prim());
44     }
45     return 0;
46 }

C/C++ (kruskal):

 1 #include <cstdio>
 2 #include <climits>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 int n, my_line, my_pre[110];
 7 struct node
 8 {
 9     int a, b, a_b_dis;
10 }my_dis[6000];
11 
12 bool cmp(node a, node b)
13 {
14     return a.a_b_dis < b.a_b_dis;
15 }
16 
17 int my_find(int x)
18 {
19     int n = x;
20     while (n != my_pre[n])
21         n = my_pre[n];
22     int i = x, j;
23     while (my_pre[i] != n)
24     {
25         j = my_pre[i];
26         my_pre[i] = n;
27         i = j;
28     }
29     return n;
30 }
31 
32 void my_join(int a, int b)
33 {
34     int n1 = my_find(a), n2 = my_find(b);
35     my_pre[n1] = n2;
36 }
37 
38 int my_kruskal()
39 {
40     /**
41         Initialize
42     */
43     int my_ans = 0;
44     sort(my_dis, my_dis + my_line, cmp);
45     for (int i = 1; i <= n; ++ i)
46         my_pre[i] = i;
47 
48     for (int i = 0; i < my_line; ++ i)
49     {
50         int a = my_find(my_dis[i].a), b = my_find(my_dis[i].b);
51         if (a != b)
52         {
53             my_pre[a] = b;
54             my_ans += my_dis[i].a_b_dis;
55         }
56     }
57 
58     return my_ans;
59 }
60 
61 int main()
62 {
63     while (scanf("%d", &n), n)
64     {
65         my_line = n * (n-1) / 2;
66         for (int i = 0; i < my_line; ++ i)
67             scanf("%d%d%d", &my_dis[i].a, &my_dis[i].b, &my_dis[i].a_b_dis);
68         printf("%d
", my_kruskal());
69     }
70     return 0;
71 }

hdu 1233 还是畅通工程 (prim， kruskal)

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