hdu 1102 Constructing Roads (kruskal)

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原文链接：http://www.cnblogs.com/GetcharZp/p/9427525.html

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本文详细解析了如何使用最小生成树算法解决村庄间道路建设问题，通过Kruskal算法实现所有村庄的连接，同时确保总道路长度最短。文章提供了完整的C/C++代码示例，包括输入村庄数量、距离矩阵及已建道路信息，通过Kruskal算法求解最小生成树，最终输出所需建设的总道路长度。

Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28154 Accepted Submission(s): 10717

Problem Description
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input
The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output
You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input
3
0 990 692
990 0 179
692 179 0
1
1 2

Sample Output
179

C/C++:

 1 #include <cstdio>
 2 #include <climits>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 int n, m, my_map[110][110], 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 (n != my_pre[i])
24     {
25         j = my_pre[i];
26         my_pre[i] = n;
27         i = j;
28     }
29     return n;
30 }
31 
32 int my_kruskal()
33 {
34     int my_cnt = 0, my_ans = 0;
35     for (int i = 1; i < n; ++ i)
36     {
37         for (int j = i + 1; j <= n; ++ j)
38         {
39             my_dis[my_cnt].a = i;
40             my_dis[my_cnt].b = j;
41             my_dis[my_cnt].a_b_dis = my_map[i][j];
42             my_cnt ++;
43         }
44     }
45     sort(my_dis, my_dis + my_cnt, cmp);
46     for (int i = 0; i < my_cnt; ++ i)
47     {
48         int n1 = my_find(my_dis[i].a), n2 = my_find(my_dis[i].b);
49         if (n1 != n2)
50         {
51             my_pre[n1] = n2;
52             my_ans += my_dis[i].a_b_dis;
53         }
54     }
55     return my_ans;
56 }
57 
58 int main()
59 {
60     /**
61         Date Input Initialize
62     */
63     while (~scanf("%d", &n))
64     {
65         for (int i = 1; i <= n; ++ i)
66             for (int j = 1; j <= n; ++ j)
67                 scanf("%d", &my_map[i][j]);
68         scanf("%d", &m);
69         for (int i = 1; i <= n; ++ i)
70             my_pre[i] = i;
71         for (int i = 0; i < m; ++ i)
72         {
73             int a, b, a_temp, b_temp;
74             scanf("%d%d", &a, &b);
75             a_temp = my_find(a), b_temp = my_find(b);
76             my_pre[a_temp] = b_temp;
77         }
78         printf("%d\n", my_kruskal());
79     }
80     return 0;
81 }