hdu 1856（并查集）More is better

More is better
Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 26832    Accepted Submission(s): 9611


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
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8
 


Sample Output
4
2


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
 
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题意：这道题的意思是一个人从所给的关系中最多能选择多少个人，并且选择的这些人，是朋友，不管直接还是间接。
看样例：
4
1 2
3 4
5 6
1 6
***
所给的这个样例中，1和2是朋友；3和4；5和6；1和6是朋友，根据所给的关系，可以知道
1,2,5,6是一个朋友圈，人数为4
---
4
1 2
3 4
5 6
7 8
***
所给的这个样例中，1和2是朋友；3和4；5和6；7和8；根据所给的关系，可以知道
这里有4个朋友圈，其中每个朋友圈里的人数都是2，也就是最多人数为2
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并查集

——---

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>using namespace std;const int maxn=10000001;int f[maxn],sum[maxn];int init()//初始化 {for(int i=1;i<=maxn;i++){f[i]=i;sum[i]=1;//开始时数量都为1，根节点为自己}}int find(int x)//找根节点 {if(f[x]!=x)f[x]=find(f[x]);return f[x];}void join(int a,int b){int t1=find(a);int t2=find(b);if(t1!=t2){f[t1]=t2;sum[t2]+=sum[t1];//合并集合中的元素个数 } }   int main(){int a,b,n,i,max;while(scanf("%d",&n)!=EOF){if(n==0){printf("1\n");continue;}max=0;init();for(i=0;i<n;i++){scanf("%d%d",&a,&b);if(a>max)max=a;if(b>max)max=b; join(a,b);}int scnt=0;for(i=1;i<=max;i++)if(sum[i]>scnt)//找朋友圈人数最多的 scnt=sum[i];printf("%d\n",scnt);}return 0;}