05-树8 File Transfer

并查集

简单并查集：输入N，代表有编号为1、2、3……N电脑。下标从1开始。初始化为-1。合并后根为负数，负数代表根，其绝对值代表个数。

We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other?

Input Specification:

Each input file contains one test case. For each test case, the first line contains $10^4), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and N. Then in the following lines, the input is given in the format:$

I c1 c2

where I stands for inputting a connection between c1 and c2; or

C c1 c2

where C stands for checking if it is possible to transfer files between c1and c2; or

where S stands for stopping this case.

Output Specification:

For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where kis the number of connected components in this network.

Sample Input 1:

Sample Output 1:

no
no
yes
There are 2 components.

Sample Input 2:

Sample Output 2:

no
no
yes
yes
The network is connected.

 1 #include <stdio.h>
 2 
 3 #define MaxN 10001                  /* 集合最大元素个数 */
 4 
 5 typedef int ElementType;           /* 默认元素可以用非负正数表示 */
 6 typedef int SetName;               /* 默认用根节点下标作为集合名称 */
 7 ElementType SetType[MaxN]; /* 假设集合元素下标从1开始 */
 8 
 9 void Union( ElementType S[], SetName Root1, SetName Root2 );
10 SetName Find( ElementType S[], ElementType X );
11 
12 int main()
13 {
14     int N;
15     int computerA,computerB;
16     scanf("%d",&N);
17     for(int i = 0; i < N+1; i++) 
18         SetType[i] = -1;
19     char operation;
20     scanf("%c",&operation);
21     while(operation != 'S') {
22         if(operation == 'I') {                    //inputting a connection
23             scanf("%d %d",&computerA, &computerB);
24             Union(SetType,Find(SetType,computerA), Find(SetType,computerB));
25         }else if(operation == 'C') {            //check
26             scanf("%d %d",&computerA, &computerB);
27             if(Find(SetType,computerA) == Find(SetType,computerB)) { //是否是同一个根 
28                 printf("yes\n");
29             }else {
30                 printf("no\n");
31             }
32         }
33         scanf("%c",&operation);
34     }
35     int components = 0;
36     for(int i = 1; i < N+1; i++) {
37         if(SetType[i] < 0)
38             components++;
39     }
40     if(components == 1)
41         printf("The network is connected.\n");
42     else
43         printf("There are %d components.\n",components);
44     return 0;
45 } 
46 /* 这里默认Root1和Root2是不同集合的根结点 */
47 void Union( ElementType S[], SetName Root1, SetName Root2 )
48 {
49     /* 保证小集合并入大集合 */
50     if ( S[Root2] < S[Root1] ) { /* 如果集合2比较大 */
51         S[Root2] += S[Root1];     /* 集合1并入集合2  */
52         S[Root1] = Root2;
53     }
54     else {                         /* 如果集合1比较大 */
55         S[Root1] += S[Root2];     /* 集合2并入集合1  */
56         S[Root2] = Root1;
57     }
58 }
59 
60 SetName Find( ElementType S[], ElementType X )
61 { /* 默认集合元素全部初始化为-1 */
62     if ( S[X] < 0 ) /* 找到集合的根 */
63         return X;
64     else
65         return S[X] = Find( S, S[X] ); /* 路径压缩 */
66 }

posted on 2016-04-04 14:32 kuotian 阅读(1014) 评论(0) 编辑收藏举报