POJ 1861 Network (模版kruskal算法)

浏览数：28 / 时间：2015年06月09日

Network
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 12563 Accepted: 4826 Special Judge
Description

Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
Input

The first line of the input contains two integer numbers: N - the number of hubs in the network (2 <= N <= 1000) and M - the number of possible hub connections (1 <= M <= 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs.
Output

Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.
Sample Input

4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1
Sample Output

1
4
1 2
1 3
2 3
3 4

View Question

代码WA了，待查找原因

 1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 #include<cstdio>
 5 #include<algorithm>
 6 using namespace std;
 7 #define MAX 1000
 8 int father[MAX], son[MAX], Min=0x3fffffff;
 9 int v, l;
10 
11 typedef struct Kruskal //存储边的信息
12 {
13     int a;
14     int b;
15     int value;
16 };
17 
18 bool cmp(const Kruskal & a, const Kruskal & b)
19 {
20     return a.value < b.value;
21 }
22 
23 int unionsearch(int x) //查找根结点+路径压缩
24 {
25     return x == father[x] ? x : unionsearch(father[x]);
26 }
27 
28 bool join(int x, int y) //合并
29 {
30     int root1, root2;
31     root1 = unionsearch(x);
32     root2 = unionsearch(y);
33     if(root1 == root2) //为环
34         return false;
35     else if(son[root1] >= son[root2])
36         {
37             father[root2] = root1;
38             son[root1] += son[root2];
39         }
40         else
41         {
42             father[root1] = root2;
43             son[root2] += son[root1];
44         }
45     return true;
46 }
47 
48 int main()
49 {
50     int ltotal;
51     int res_f[100],res_b[100];
52     Kruskal edge[MAX];
53     while(scanf("%d%d",&v,&l)!=EOF)
54     {
55         ltotal = 0;
56         for(int i = 1; i <= v; ++i) //初始化
57         {
58             father[i] = i;
59             son[i] = 1;
60         }
61         for(int i = 1; i <= l ; ++i)
62         {
63             scanf("%d%d%d", &edge[i].a, &edge[i].b, &edge[i].value);
64         }
65         sort(edge + 1, edge + 1 + l, cmp); //按权值由小到大排序
66         for(int i = 1; i <= l; ++i)
67         {
68             if(join(edge[i].a, edge[i].b))
69             {
70                 res_f[ltotal]=edge[i].a; res_b[ltotal]=edge[i].b;
71                 ltotal++; //边数加1
72                 //cout<<edge[i].a<<" "<<edge[i].b<<endl;
73                 if(edge[i].value < Min)
74                     Min=edge[i].value;
75             }
76         }
77         printf("%d\n%d\n",Min,ltotal);
78         for(int i=0;i<ltotal;i++){
79             printf("%d %d\n",res_f[i],res_b[i]);
80         }
81     }
82     return 0;
83 }