九度 題目1444：More is better

九度 題目1444：More is better

原題OJ鏈接：http://ac.jobdu.com/problem.php?pid=1444

題目描述：

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.

輸入：

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)

輸出：

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

樣例輸入：

4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8

樣例輸出：

4
2

解題思路：

輸出並查集中元素個數的最大值；
爲了計算每個並查集的元素個數，在每個集合的樹根結點記錄該集合所包含元素的個數，用數組root[ ] 來記錄，在合併時累加被合併兩個集合包含的元素個數。

源代碼：

#include<iostream>
#include<cstring>
using namespace std;
#define MAX_N 10000005
int Tree[MAX_N];
int root[MAX_N];

int findRoot(int x){
    if(Tree[x]==-1) return x;
    else{
        int tmp=findRoot(Tree[x]);
        Tree[x]=tmp;
        return tmp;
    }
}

int main(){
    int n,A,B;
    while(cin>>n){
        for(int i=0;i<MAX_N;i++){
            Tree[i]=-1;
            root[i]=1;
            /*用root[i]表示以結點i爲根的樹的結點個數，
            其中保存數據僅當Tree[i]爲-1時 
            即該結點爲樹的根結點時有效*/
        }
        for(int i=0;i<n;i++){
            cin>>A>>B;
            A=findRoot(A);
            B=findRoot(B);
            if(A!=B){
                Tree[A]=B;
                root[B]+=root[A];
                /*合併集合時，將子樹的根結點上保存的該集合元素
                個數累加到合併後新樹的樹根上
                */
            } 
        }

        int count=1;
        for(int i=0;i<MAX_N;i++){
            if(root[i]>count && Tree[i]==-1){
                count=root[i];
            }
        }
        cout<<count<<endl;
    }
    return 0;
}