hdu 1856 More is better (并查集)

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 11952    Accepted Submission(s): 4417

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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//359MS    78584K    889 B    G++
/*

    题意：
        给出n对朋友关系，求直接或间接认识的最大的集体
        
    并查集：
        赤裸裸的并查集，直接并查集解法，注意数组大小和没朋友关系的情况。 

*/ 
#include<stdio.h>
#include<string.h>
int set[10000005];
int num[10000005];
int find(int x)
{
    if(set[x]==x) return x;
    return find(set[x]);
}
void merge(int a,int b)
{
    int x=find(a);
    int y=find(b);
    if(x>y) set[x]=y;
    else set[y]=x;
}
int main(void)
{
    int n,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        int m=0;
        for(int i=0;i<10000005;i++) set[i]=i;
        for(int i=0;i<n;i++){
            scanf("%d%d",&a,&b);
            merge(a,b);
            if(a>m) m=a;
            if(b>m) m=b;
        }
        memset(num,0,sizeof(num));
        int cnt=0;
        for(int i=1;i<=m;i++){
            int t=find(set[i]);
            num[t]++;
            if(num[t] > cnt) cnt=num[t];
        } 
        if(cnt==0) printf("1
"); //没朋友关系时只能留一个 
        else printf("%d
",cnt);    
    }
    return 0;
}