模板--并查集

问题描述：
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.

输入：

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) .
输出：

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.
样例输入：
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8
样例输出：
4
2

思路：
利用并查集的基础，进行修改，注意数组的范围。以及路径压缩。

代码：

#include<iostream>#include<cstring>#include<stdio.h>const int MX=1000000;using namespace std;int pre[MX],t[MX];int fid(int x){    int r=x,t;    while(r!=pre[r])    {        r=pre[r];    }     while (x != r)        { t = pre[x];        pre[x] = r;     x = t; }    return x;}void mix(int a,int b){    int x=fid(a),y=fid(b);    if(x!=y)    {        pre[x]=y;    }}int main(){    int n;   while(scanf("%d",&n)!=EOF)    {    int maxn=0;       for(int i=1;i<=MX;i++)         {pre[i]=i;         t[i]=0;}       for(int i=1;i<=n;i++)        {           int a,b;           cin>>a>>b;           if(a>maxn)maxn=a;           if(b>maxn)maxn=b;           if(fid(a)!=fid(b))           mix(a,b);        }     int ct=1;       for(int i=1;i<=maxn;i++)       {        t[fid(i)]++;       }      for(int i=1;i<=maxn;i++)       {              if(t[i]>ct)              ct=t[i];      }    cout<<ct<<endl;   } return 0;}