HDOJ 1856 More is better（并查集）

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

题目大意：房间里有10000000个男孩，编号从1到10000000。一共有n次操作，每次操作指定2个男孩成为朋友（直接的或间接的），问有朋友关系的男孩最多有几个。

解题思路：利用并查集维护男孩之间的关系，rank[i]保存以i为根节点的集合内有多少男孩。注意n=0时输出1。

代码如下：

#include 
const int maxn = 100005;
const int maxm = 10000005;
int par[maxm],rank[maxm];
int a[maxn],b[maxn];
void init(int n)
{
	for(int i = 1;i <= n;i++){
		par[i] = i;
		rank[i] = 1;
	}
}

int find(int x)
{
	return par[x] == x ? x : par[x] = find(par[x]);
}

void unite(int x,int y)
{
	x = find(x);
	y = find(y);
	if(x == y)
		return ;
	par[y] = x;
	rank[x] += rank[y];
	rank[y] = 1;
}

bool same(int x,int y)
{
	return find(x) == find(y);
}

int main()
{
	int n;
	while(scanf("%d",&n) != EOF){
		int mx = 1;
		for(int i = 0;i < n;i++){
			scanf("%d %d",&a[i],&b[i]);
			mx = std::max(mx,std::max(a[i],b[i]));
		}
		init(mx);
		for(int i = 0;i < n;i++){
			unite(a[i],b[i]);
		}
		int ans = 0;
		for(int i = 1;i <= mx;i++){
			ans = std::max(rank[i],ans);
		}
		printf("%d\n",ans);
	}
	return 0;
}