hdu-1856 More is better

最新推荐文章于 2019-06-29 14:23:40 发布

B_mess

最新推荐文章于 2019-06-29 14:23:40 发布

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分类专栏：并查集

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本文介绍了一个算法问题，即如何计算在一个包含大量人员的房间中，根据直接朋友关系能够形成的最大朋友圈规模。通过输入一系列直接朋友对，算法需找出能够构成的最大朋友圈包含的人数。

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More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 18778 Accepted Submission(s): 6902

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

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

#include<stdio.h>  
#define N 10000000  

int father[N],num[N];  

void initial()
{  
    int i;  
    for(i=1;i<=N;i++)  
    {  
        father[i]=i;  
        num[i]=1;
    }  
}  

int find(int x) 
{  
    if(father[x]!=x)  
        father[x]=find(father[x]);  
    return father[x];  
}  

void hebing(int a,int b) 
{  
    int p=find(a);  
    int q=find(b);  
    if(p!=q)  
    {  
        father[p]=q;  
        num[q]+=num[p];
    }  
} 
 
int main()  
{  
    int n,a,b,i,sum,max;  
    while(~scanf("%d",&n))  
    {  
        if(n==0)  
        {  
            printf("1\n");  
            continue;  
        }  
        max=0;  
        initial(); 
        for(i=0;i<n;i++)  
        {  
            scanf("%d%d",&a,&b);  
            if(a>max)  
                max=a;  
            if(b>max)  
                max=b;  
            hebing(a,b);  
        }  
        int Max=0;  
        for(i=1;i<=max;i++)  
            if(num[i]>Max) 
                Max=num[i];  
        printf("%d\n",Max);  
    }  
    return 0;  
}