HDU 1856 More is better

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
     
     
     
     
     
     
     
     

      
      
      
      这个题用rank数组记录当前点的孩子有多少个：
     
     
     
     
     
     
     
     

      
      
      
      合并时，判断两个节点的rank值哪个大，rank值大的作为父亲节点；
     
     
     
     
     
     
     
     

      
      
      
      language:c++
     
     
     
     
     
     
     
     

      
      
      
      code:
     
     
     
     
     
     
     
     

      
      
      
      
#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>

using namespace std;

struct node
{
    int x,y;
    node(int a,int b):x(a),y(b) {}
    //赋值构造
};

int father[10000002],rank[10000002];

void set(int x)
{
    father[x]=x;
    rank[x]=1;
    //初始化为1，记录当前节点加孩子的个数。
}

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

bool Union(int x,int y)
{
    int root1=find(x);
    int root2=find(y);
    if(root1!=root2)
    {
        //rank大的作根，且根的rank值加上小的rank值。
        if(rank[root1]<rank[root2])
        {
            father[root1]=root2;
            rank[root2]+=rank[root1];
        }
        else
        {
            rank[root1]+=rank[root2];
            father[root2]=root1;
        }
        return true;
    }
    return false;
}



int main()
{
    //freopen("in.txt","r",stdin);
    int pair;
    vector<node>vect;
    while(scanf("%d",&pair)!=EOF)
    {
        if(pair==0)
        {//当数据为0组的时候要输出1；
            printf("1\n");
            continue;
        }
        vect.clear();
        int max=0;
        while(pair--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            set(u),set(v);//设初始值
            vect.push_back(node(u,v));//记录输入的值
        }
        for(int i=0; i<vect.size(); i++)
        {
            //一对一对地合并
            Union(vect[i].x,vect[i].y);
        }
        for(int i=0; i<vect.size(); i++)
        {
            //找出rank值最大的
            if(max<rank[vect[i].x])
                max=rank[vect[i].x];
            if(max<rank[vect[i].y])
                max=rank[vect[i].y];
        }
        printf("%d\n",max);
    }
    return 0;
}