并查集练习 1107 Social Clusters (30分)

关于有关并查集的题主要应掌握两点：
father[]数组：记录每个结点的父结点
findAncestor():找到该结点所在集合的根结点

1107 Social Clusters (30分)

该题中还需要求出每个集合中结点的个数，因此只需设置一个isAncestor[]数组，isAncestor[i]表示以i作为根结点的集合中有多少结点，若i不是根结点，则isAncestor[i]=0。

for (int i = 1; i <= n; i++)
	isAncestor[findAncestor(i)]++;

const int maxn = 1010;
int lastPerson[maxn] = { 0 };  //记录当前喜欢某活动的最后一个人
int father[maxn];
int isAncestor[maxn] = { 0 };

int findAncestor(int x)
{
	if (father[x] == x)
		return x;
	else
		return findAncestor(father[x]);
}
bool cmp(int a, int b)
{
	return a > b;
}
int main()
{
	int n, act,num;
	scanf("%d", &n);
	//人从1开始编号,初始化father数组
	for (int i = 1; i <= n; i++)
		father[i] = i;
	for (int i = 1; i <= n; i++)
	{
		scanf("%d: ", &num);
		for (int j = 0; j < num; j++)
		{
			scanf("%d", &act);
			//已经有人喜欢该活动
			if (lastPerson[act] != 0)
			{
				int lp = lastPerson[act];
				int anc1 = findAncestor(i);
				int anc2 = findAncestor(lp);
				//若这两个结点还没有在同一个集合中，合并
				if (anc1 != anc2)
					father[anc1] = anc2;
			}
			lastPerson[act] = i;
		}
	}
	int uNum = 0;
	for (int i = 1; i <= n; i++)
		isAncestor[findAncestor(i)]++;
	for (int i = 0; i < maxn; i++)
		if (isAncestor[i] != 0)
			uNum++;
	printf("%d\n", uNum);
	sort(isAncestor, isAncestor + maxn, cmp);
	for (int i = 0; i < uNum-1; i++)
		printf("%d ", isAncestor[i]);
     printf("%d", isAncestor[uNum-1]);   
}