HDU1856:More is better

点击打开题目链接

More is better Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others) Total Submission(s): 9760    Accepted Submission(s): 3592 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 Hint A 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.   Author lxlcrystal@TJU   Source HDU 2007 Programming Contest - Final   Recommend lcy

=====================================题目大意=====================================

Mr Wang需要尽可能多的男孩来帮助他完成一个复杂的项目，不过Mr Wang选择的男孩必须相互之间都是朋友（直接的或间接的）。

给出前来应聘的男孩之间的直接朋友关系，求解Mr Wang所能选择的男孩的最大人数。

=====================================算法分析=====================================

并查集。

=======================================代码=======================================

#include<stdio.h>
#include<string.h>

const int MAXN=10000005;

int N,M,P[MAXN],S[MAXN];

int Find(int x) 
{
	return !P[x]?x:P[x]=Find(P[x]);
}

int main()
{
    while(scanf("%d",&N)==1)
    {
        memset(P,0,sizeof(P));
        memset(S,0,sizeof(S));
		int A,B,Ans=0;
        while(N--)
        {
            scanf("%d%d",&A,&B);
			int PA=Find(A);
            int PB=Find(B);
            if(PA!=PB)
            {
                P[PB]=PA;                             //若该语句更改为P[PA]=PB;
                if((S[PA]+=S[PB]+1)>Ans) Ans=S[PA];   //则该语句需改为if((S[PB]+=S[PA]+1)>Ans) Ans=S[PB];
            }
        }
        printf("%d\n",Ans+1);
    }
    return 0;
}