More is better(並差集)

More is better

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
核心代碼：在並差集的合併中改代碼，找一棵節點最多的樹

void mix(int a,int b)
{
	int fx=find(a);
	int fy=find(b);
	if(fx!=fy)
	{
	  pre[fx]=fy;
	  mark[fy]+=mark[fx];//記錄每棵樹的節點數 
	 if(mark[fy]>countt)
	 countt=mark[fy];	
	}	
}

例如：在給的第一組數據中
1 2 3 4 5 6 1 6

#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 10000005
int pre[N],mark[N],countt;
using namespace std;
int find(int x)
{
	int r=x;
	while(pre[r]!=r)
	r=pre[r];
	
	int i=x,j;
	while(i!=r)
	{
	 j=pre[i];
	 pre[i]=r;
	 i=j;
	}
	return r;
}

void init()
{
	for(int i=0;i<=N;i++)
	{
		pre[i]=i;
		mark[i]=1;
	}
}

void mix(int a,int b)
{
	int fx=find(a);
	int fy=find(b);
	if(fx!=fy)
	{
	  pre[fx]=fy;
	  mark[fy]+=mark[fx];//記錄每棵樹的節點數
	 
	 if(mark[fy]>countt)
	 countt=mark[fy];	
	}	
}

int main()
{
	int t,p1,p2;
	while(scanf("%d",&t)!=EOF)
	{
		countt=1;
		init();
		while(t--)
		{
			scanf("%d %d",&p1,&p2);
			mix(p1,p2);
		 }
	     printf("%d\n",countt);
	}
	return 0;
}

我本來想用map寫 嘻嘻但是存儲超限 而且運行太慢

#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 10000005
int pre[N],mark[N],countt;
#include <map>
using namespace std;
map <int,int>mp;
int find(int x)
{
    int r=x;
    while(pre[r]!=r)
    r=pre[r];
    
    int i=x,j;
    while(i!=r)
    {
     j=pre[i];
     pre[i]=r;
     i=j;
    }
    return r;
}

void init()
{
    for(int i=0;i<=N;i++)
    {
        pre[i]=i;
        mark[i]=0;
    }
}

void mix(int a,int b)
{
    int fx=find(a);
    int fy=find(b);
    if(fx!=fy)
    {
      pre[fx]=fy;
      //mark[fy]+=mark[fx];//記錄每棵樹的節點數
     //mp[fy]++;
     // if(mp[fy]>countt)
     // countt=mp[fy];    
    }    
}

int main()
{
    int t,p1,p2;
    while(scanf("%d",&t)!=EOF)
    {
        if(t==0)
        {
        printf("1\n");
        continue;
        }
        countt=0;
        init();
        while(t--)
        {
            scanf("%d %d",&p1,&p2);
            mix(p1,p2);
            mark[p1]=1,mark[p2]=1;
            
         }
        for(int i=0;i<=N;i++)
        {
            if(mark[i])
            mp[find(i)]++;
        }
         for(int i=0;i<=N;i++)
         {
         if(mp[i]>countt)
         countt=mp[i];
         } 
         printf("%d\n",countt);
         mp.clear(); 
    }
    return 0;
}