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[110][110];

int my_prim()
{
    int my_pos = 1, my_book[110] = {0, 1}, my_dis[110] = {0, INT_MAX}, my_ans = 0;
    for (int i = 2; i <= n; ++ i)
        my_dis[i] = my_map[my_pos][i];
    for (int i = 1; i < n; ++ i)
    {
        int my_temp = INT_MAX;
        for (int j = 1; 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] = 1;
        for (int j = 1; 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-1) / 2;
        for (int i = 0; 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
", my_prim());
    }
    return 0;
}

 C/C++ (kruskal):

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

int n, my_line, my_pre[110];
struct node
{
    int a, b, a_b_dis;
}my_dis[6000];

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 = 0;
    sort(my_dis, my_dis + my_line, cmp);
    for (int i = 1; i <= n; ++ i)
        my_pre[i] = i;

    for (int i = 0; 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-1) / 2;
        for (int i = 0; i < my_line; ++ i)
            scanf("%d%d%d", &my_dis[i].a, &my_dis[i].b, &my_dis[i].a_b_dis);
        printf("%d
", my_kruskal());
    }
    return 0;
}