HDOJ 1856 More is better （并查集）

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)

Total Submission(s): 19234    Accepted Submission(s): 7075

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.

注 - 此题为：  HDOJ 1856 More is better  （并查集）

说明： 求 所有 根节点中 ，集合中元素最多的个数.

第一个 例子： {1,2,5,6}  {3,4}      最多的是           4；

第二个例子： {1,2} {3,4} {5,6} {7,8}   最多的是    2。

注意：    压缩路径  否则超时，不能AC    ；   N为 0 ，要输出 1 ,因为集合中最少有一个元素（自身），

已AC代码：

#include
#define M 10000000
#define max(x,y) (x > y ? x : y)
int num[M+10],per[M+10];

void itoa()
{
	for(int i = 1; i < M+10; ++i)
	{
		per[i] = i;
		num[i] = 1;     //初始每个集合中有一个元素 
	}
}

int find(int x)    //压缩路径  否则超时，不能AC 
{           //把 一个集合中所有节点 连到 根节点上 
	return x == per[x] ? x : per[x] = find(per[x]);
}

void join(int a,int b)
{
	int fa = find(a);
	int fb = find(b);
	if(fa != fb)
	{
		per[fa] = fb;
		num[fb] += num[fa];    // 计算合并后的元素个数 
	}
}

int main()
{
	int n,i,a,b,m;
	while(scanf("%d",&n)!=EOF)
	{
		if(n == 0)   // n 为 0 ，要输出 1 
		{
			printf("1\n");
			continue;
		}
		
		itoa();   //初始化 
		
		m=0;
		for(i = 0; i < n; ++i)
		{
			scanf("%d%d",&a,&b);
			join(a,b); // 加入 集合 
			m = max(m,max(a,b));    // 最大的编号 
		}
		int MAX = 0;
		for(i = 1;i <= m; ++i)  // 元素最多的集合 
			if(num[i] > MAX)
				MAX = num[i];
		printf("%d\n",MAX);
	}
	return 0;
}