hdu 1856 More is better （并查集）

问题描述：
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

大致题意：
王老师要找一些男生帮助他完成一项工程。要求最后挑选出的男生之间都是朋友关系，可以说直接的，也可以是间接地。问最多可以挑选出几个男生（最少挑一个)。

思路分析：
简单并查集。
建个数组，记录个数就行了。合并的时候要将个数加过去。
还要注意的是 输入0的时候输出1.

ac代码：

#include<bits/stdc++.h>using namespace std;int fa[10000002];int num[10000002];#define N 10000000void init()//初始化{    int i;    for(i=1;i<=N;i++)    {        fa[i]=i;        num[i]=1;/*开始时数量都为1，根节点为自己*/    }}int findd(int x){    if(x!=fa[x])        fa[x]=findd(fa[x]);    return fa[x];}void Merge(int x,int y){    int p1,p2;    p1=findd(x);    p2=findd(y);    if(p1==p2)        return ;    fa[p2]=fa[p1];    num[p1]+=num[p2];}int main(){    ios::sync_with_stdio(false);    int T,n,m;    int i,j,k;    int maxx;    maxx=-1;    while(cin>>T)    {        init();        if(T==0)        {            cout<<1<<endl;            continue;        }        while(T--)        {            cin>>n>>m;            maxx=max(maxx,max(n,m));            Merge(n,m);        }        int sum=-1;        for(i=1;i<=maxx;i++)        {            sum=max(sum,num[i]);        }        cout<<sum<<endl;    }    return 0;}