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。省政府“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)：

 #include <cstdio>
 #include <climits>
 using namespace std;

 int n, my_map[][];

 int my_prim()
 {
     int my_pos = , my_book[] = {, }, my_dis[] = {, INT_MAX}, my_ans = ;
     for (int i = ; i <= n; ++ i)
         my_dis[i] = my_map[my_pos][i];
     for (int i = ; i < n; ++ i)
     {
         int my_temp = INT_MAX;
         for (int j = ; j <= n; ++ j)
         {
             if (!my_book[j] && my_dis[j] < my_temp)
             {
                 my_temp = my_dis[j];
                 my_pos = j;
             }
         }
         my_ans += my_temp;
         my_book[my_pos] = ;
         for (int j = ; j <= n; ++ j)
             if (my_dis[j] > my_map[j][my_pos])
                 my_dis[j] = my_map[j][my_pos];
     }
     return my_ans;
 }

 int main()
 {
     while (scanf("%d", &n), n)
     {
         int my_line = n * (n-) / ;
         for (int i = ; i < my_line; ++ i)
         {
             int a, b, a_b_dis;
             scanf("%d%d%d", &a, &b, &a_b_dis);
             my_map[a][b] = my_map[b][a] = a_b_dis;
         }
         printf("%d\n", my_prim());
     }
     return ;
 }

C/C++ (kruskal):

 #include <cstdio>
 #include <climits>
 #include <algorithm>
 using namespace std;

 int n, my_line, my_pre[];
 struct node
 {
     int a, b, a_b_dis;
 }my_dis[];

 bool cmp(node a, node b)
 {
     return a.a_b_dis < b.a_b_dis;
 }

 int my_find(int x)
 {
     int n = x;
     while (n != my_pre[n])
         n = my_pre[n];
     int i = x, j;
     while (my_pre[i] != n)
     {
         j = my_pre[i];
         my_pre[i] = n;
         i = j;
     }
     return n;
 }

 void my_join(int a, int b)
 {
     int n1 = my_find(a), n2 = my_find(b);
     my_pre[n1] = n2;
 }

 int my_kruskal()
 {
     /**
         Initialize
     */
     int my_ans = ;
     sort(my_dis, my_dis + my_line, cmp);
     for (int i = ; i <= n; ++ i)
         my_pre[i] = i;

     for (int i = ; i < my_line; ++ i)
     {
         int a = my_find(my_dis[i].a), b = my_find(my_dis[i].b);
         if (a != b)
         {
             my_pre[a] = b;
             my_ans += my_dis[i].a_b_dis;
         }
     }

     return my_ans;
 }

 int main()
 {
     while (scanf("%d", &n), n)
     {
         my_line = n * (n-) / ;
         for (int i = ; i < my_line; ++ i)
             scanf("%d%d%d", &my_dis[i].a, &my_dis[i].b, &my_dis[i].a_b_dis);
         printf("%d\n", my_kruskal());
     }
     return ;
 }