hdu 1856 More is better

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

此题是并查集的简单应用。

#include
int father[10000001],rank[10000001];//father[x]记录x的父节点，rank[x]保存根节点数目
int find(int x)//寻找x的父节点
{
    if(x!=father[x])
        father[x]=find(father[x]);//回溯时路径压缩
    return father[x];
}
void uion(int x,int y)//合并x，y所在的集合
{
    x=find(x);
    y=find(y);
    if(x==y) return ;
    if(rank[x]>rank[y])
    {
        father[y]=x;
        rank[x]+=rank[y];
    }
    else if(rank[x]<rank[y])
    {
        father[x]=y;
        rank[y]+=rank[x];
    }
    else {
        rank[x]*=2;
        father[y]=x;
    }
}
int main()
{
    int n,i,j,max,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        max=1;
        for(i=0;i<10000000;i++)//初始化
        {
            father[i]=i;
            rank[i]=1;
        }
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a,&b);
            uion(a,b);
        }
        max=rank[0];
        for(i=1;i<10000000;i++)//寻找最大的集合
            if(max<rank[i])
                max=rank[i];
        printf("%d\n",max);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/duan-to-success/p/3506057.html