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.

 #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 ;

 }

 #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 ;

 }