HDU 1856 More is better (and query set)

HDU 1856 More is better (and query set)

More is better Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission (s): 15639 Accepted Submission (s): 5761


Problem DescriptionMr Wang wants some boys to help him with a project. Because the project is rather complex, The more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. the boy who are not been chosen has to leave the room immediately. there are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. after Mr Wang's selection any two of them who are still in this room shoshould be friends (direct or indirect), or there is only one boy left. given all the direct friend-pairs, you shoshould decide the best way.

InputThe first line of the input contains an integer n (0 ≤ n ≤0 ≤ 100 000)-the number of direct friend-pairs. the following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≤ B, 1 ≤ A, B ≤ 10000000)
OutputThe output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input

41 23 45 61 641 23 45 67 8

Sample Output

42HintA and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result.In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.  

This is also a query question, asking the maximum number of friends, in fact, is to calculate the maximum value of ranks.

# Include
 
  
# Include
  
   
Using namespace std; # define M 10000010int ranks [M]; // At the beginning, I thought there was not enough space. As a result, the space given by this question is very large int pre [M]; int _ find (int node) {int t = node; if (node = pre [node]) return node; else return pre [node] = _ find (pre [node]);} int _ union (int a, int B) {int p = _ find (); int q = _ find (B); if (p! = Q) {if (ranks [p]
   
    
Max) max = ranks [I];} printf ("% d \ n", max);} return 0 ;}