Hdoj 1272 problem-solving Report-query set

Xiaoxi's maze Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/others) total submission (s): 11043 accepted submission (s): 3261Problem description the last time gardon's Labyrinth Castle was played for a long time (see Problem B). Now she wants to design a maze for gardon. However, she had different ideas for designing the maze. First, she thought that all the channels should be connected in two directions. That is to say, if one channel connects room A and B, it can be used to go from room A to Room B, or from Room B to room A to improve the difficulty, john hopes that any two rooms can have only one path to communicate with each other (unless they go back ). Xiaoxi now gives you her design drawing to help you determine whether her design drawing conforms to her design idea. For example, in the following example, the first two conditions are met, but the last one has two methods from 5 to 8.
The input contains multiple groups of data. Each group of data is a list of integer pairs ending with 0, indicating the numbers of the two rooms connected by one channel. The number of the room must be at least 1 and cannot exceed 100000. There is an empty row between each two groups of data.
The entire file ends with two-1 characters. For each group of input data, output only contains one row. If the maze conforms to Xiao Xi's idea, "yes" is output; otherwise, "no" is output ". Sample input6 8 5 3 5 2 6 4 5 6 0 0 8 1 7 3 6 2 8 9 7 5 7 7 8 6 0 0 3 8 6 8 6 4 5 3 5 6 5 2 0 0-1-1 sample outputyes Yes No authorgardon sourcehdu 2006-4 Programming Contest recommendlxj

1 // The stack cannot be recursively called when the compression path is created.
2 // when 0 is input, yes is output.
3
4 # include <cstdio>
5 # define N 100100 + 10
6
7 int father [N];
8 int number [N];
9
10 void ini (void)
11 {
12 for (int I = 1; I <N; I ++)
13 {
14 number [I] = 0;
15 father [I] = I;
16}
17}
18
19 int Find_Set (int)
20 {
21 int B =;
22 while (father [B]! = B)
23 B = father [B];
24 father [a] = B;
25 return father [a];
26}
27
28 void Union (int a, int B)
29 {
30 a = Find_Set ();
31 B = Find_Set (B );
32 father [a] = B;
33}
34
35 int main ()
36 {
37 // freopen ("input.txt", "r", stdin );
38 int X, Y, num, Max, I, num2;
39 while (scanf ("% d", & X, & Y) = 2 & x> = 0)
40 {
41 num = 0;
42 max = 0;
43 num2 = 0;
44 INI ();
45 while (X! = 0 | y! = 0)
46 {
47 num ++;
48 number [x] = 1;
49 number [y] = 1;
50 Union (x, y );
51 scanf ("% d", & X, & Y );
52}
53 For (I = 1; I <n; I ++)
54 {
55 if (number [I] = 1)
56 Max ++;
57 if (father [I] = I & number [I] = 1)
58 num2 ++;
59}
60 // printf ("num2 = % d, num = % d, max = % d \ n", num2, num, max );
61 if (max = 0)
62 {
63 puts ("Yes ");
64 continue;
65}
66 if (num2 = 1 & num = max-1)
67 puts ("Yes ");
68 else
69 puts ("No ");
70}
71 return 0;
72}