HDU 1856 More is better 并查集

More is betterTime Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)Total Submission(s): 23649    Accepted Submission(s): 8516Problem DescriptionMr 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.InputThe 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)OutputThe output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep. Sample Input41 23 45 61 641 23 45 67 8Sample Output42HintA and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result.In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;#define MAXN 100000*2int Par[MAXN], Rank[MAXN];void Init(int n){    for (int i = 1;i <= n;i++)    {        Par[i] = i;        Rank[i] = 0;    }}int Find(int x){    if (Par[x] == x) return x;    else return Par[x] = Find(Par[x]); ///同时压缩路径}void Unite(int x, int y){    x = Find(x);y = Find(y);    if (x == y) return;    if (Rank[x]<Rank[y]) Par[x] = y;    else Par[y] = x;    if (Rank[x] == Rank[y]) Rank[x]++;}int main(void){    //freopen("F:\\test.txt","r",stdin);    int n;    while(~scanf("%d",&n))    {        Init(100000);        for(int i=1,a,b;i<=n&&scanf("%d %d",&a,&b);i++)            Unite(a,b);        for(int i=1;i<=100000;i++)  Par[i] = Find(i);//全部更新一下祖先 sort(Par+1,Par+1+100000);  int Max = 1,Count = 1,Last = Find(1);        for(int i=2;i<=100000;i++)        {            int now = Par[i];            if(now == Last) Count ++;            else            {                Max = max(Count,Max);                Last = now;                Count = 1;            }        }        Max = max(Count,Max);   //最后一组数据处理//判定最大的朋友圈        printf("%d\n",Max);    }}

来源： http://acm.hdu.edu.cn/showproblem.php?pid=1856