HDU 1856 并查集

http://acm.hdu.edu.cn/showproblem.php?pid=1856

  

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 29843    Accepted Submission(s): 10605

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

   题意没大看懂，直接并查集注意N==0时候输出1，判断最大的联通快的个数。一开始忘了路径压缩T了真是zz

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
#define pii pair<int,int>
#define inf 0x3f3f3f3f
int f[100005],tot[100005];
int getf(int v){return f[v]==v?v:f[v]=getf(f[v]);}
map<int,int>M;
int main()
{
    int N,i,j,k;
    while(scanf("%d",&N)==1){memset(tot,0,sizeof(tot));
    if(N==0){puts("1");continue;}
        M.clear();
        int p=0,ans=0,u,v;
        for(i=1;i<=100000;++i)f[i]=i;
        for(i=1;i<=N;++i)
        {
            scanf("%d%d",&u,&v);
            int _u=M[u];
            int _v=M[v];
            if(!_u){M[u]=++p;_u=p;}
            if(!_v){M[v]=++p;_v=p;}
            int fu=getf(_u);
            int fv=getf(_v);
            if(fu!=fv){
                f[fv]=fu;
            }
        }
        for(i=1;i<=p;++i) tot[getf(i)]++;
        for(i=1;i<=p;++i) ans=max(ans,tot[i]);
        printf("%d
",ans);
    }
    return 0;
}