More is better

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 19855    Accepted Submission(s): 7289

Problem Description

Mr 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 should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

 

Input

The first line of the input contains an integer n (0 ≤ n ≤ 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)

 

Output

The 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

42
Hint
A 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. 
 

 

Author

lxlcrystal@TJU

 

Source

HDU 2007 Programming Contest - Final

 

题目大意：先输入一个n，然后输入n组数，判断最大的朋友圈里面有几个人，规则是若A与B是朋友，B与C是朋友，则A与C也是朋友；

#include<iostream>#define Size 10000001using namespace std;int father[Size],a[Size],max1;int find(int x){while(x!=father[x])x=father[x];return x;}void merage(int x,int y){x=find(x);y=find(y);if(x==y)return;if(a[x]>=a[y]){father[y]=x;a[x]=a[x]+a[y];if(max1<a[x])max1=a[x];}else {father[x]=y;a[y]=a[x]+a[y];if(max1<a[y])max1=a[y];}}int main(){int i,n,x,y;while(cin>>n){if(n==0){cout<<1<<endl;continue;}max1=0;for(i=0;i<n;i++){father[i]=i;a[i]=1;}for(i=0;i<n;i++){cin>>x>>y;merage(x,y);}cout<<max1<<endl;}}

最近几天水的题够多了，该写点难度高点的题了--_--