HDOJ 1856 More is better

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 24215    Accepted Submission(s): 8695

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

 

再说点题外话，原来写的博客根本没法看!!!当时题看懂了A了觉得题好简单就不写题解了……现在我想看一下当时写了点什么都要再看一遍题!!!以后英语题一定要在底下贴翻译= =

题意：一个老师需要学生帮他干个活，为了提高效率，他要找相互认识的学生。第一行表示有N组学生相互认识，下面N行是相互认识的学生的编号。问：最多能找多少学生过来？

题解：也是求并查集的集合中元素个数的问题，不过这回是找出元素个数最大的。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int per[10000000+10],ans[10000000+10];#define N 10000000bool cmp(int a,int b){return a>b;}int Find(int x) {      if(per[x]!=x)          per[x]=Find(per[x]);      return per[x];  }  void Merge(int x,int y){int fx=Find(x);int fy=Find(y);if(fx!=fy){per[fx]=fy;ans[fy]+=ans[fx];}}struct edge{int s,e;}a[1000000+10];int main(){int i,n;while(~scanf("%d",&n)){int m=0;if(n==0){printf("1\n");continue;}for(i=0;i<n;i++){scanf("%d%d",&a[i].s,&a[i].e);if(a[i].s>m)m=a[i].s;if(a[i].e>m)m=a[i].e;}ans[0]=0;for(i=1;i<=m;i++){per[i]=i;ans[i]=1;}for(i=0;i<n;i++)Merge(a[i].s,a[i].e);int maxx=0;for(i=1;i<=m;i++)maxx=max(maxx,ans[i]);printf("%d\n",maxx);}return 0;}