poj 2985(并查集+线段树求K大数)

解题思路：这道题并查集很容易，合并时找到父节点就直接加上去就ok了。关键是如何求K大数，我一直在想用线段树怎么写，一开始想如果直接记录数的大小那肯定是没戏了，借鉴了一下别人的思路：区间[a,b]记录的是所有的数里面，等于a，a+1，a+2，......，b-1，b的个数。看到这里就应该明白了，这里线段树的用法是把它看做是一个1-n的数轴。到时候要修改某一个数，就直接在数轴上修改它。至于记录[a,b]的个数，是为了找到K大数。

参考博客：http://blog.sina.com.cn/s/blog_6fd8e0fe0100v89n.html

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 200005;
struct Segment
{
	int l,r;
	int sum;
}tree[maxn<<2];
int n,m,fa[maxn],tot[maxn];

int find(int x)
{
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}

void build(int rt,int l,int r)
{
	tree[rt].l = l, tree[rt].r = r;
	if(tree[rt].l == 1) tree[rt].sum = n;
	else tree[rt].sum = 0;
	if(l == r) return;
	int mid = (l + r) >> 1;
	build(rt<<1,l,mid);
	build(rt<<1|1,mid+1,r);
}

void update(int rt,int pos,int val)
{
	tree[rt].sum += val;
	if(tree[rt].l == tree[rt].r) return;
	int mid = (tree[rt].l + tree[rt].r) >> 1;
	if(pos <= mid) update(rt<<1,pos,val);
	else update(rt<<1|1,pos,val);
}

int query(int rt,int k)
{
	if(tree[rt].l == tree[rt].r) return tree[rt].l;
	int mid = (tree[rt].l + tree[rt].r) >> 1;
	if(tree[rt<<1|1].sum >= k) return query(rt<<1|1,k);
	else return query(rt<<1,k - tree[rt<<1|1].sum);
}

int main()
{
	int op,a,b,x,y;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i = 1; i <= n; i++)
			fa[i] = i, tot[i] = 1;
		build(1,1,n);
		for(int i = 1; i <= m; i++)
		{
			scanf("%d",&op);
			if(op)
			{
				scanf("%d",&a);
				printf("%d\n",query(1,a));
			}
			else
			{
				scanf("%d%d",&a,&b);
				x = find(a);
				y = find(b);
				if(x == y) continue;
				fa[y] = x;
				update(1,tot[x],-1);
				update(1,tot[y],-1);
				update(1,tot[x]+tot[y],1);
				tot[x] += tot[y];
				tot[y] = 0;
			}
		}
	}
	return 0;
}

用树状数组求k大数：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 200005;
struct Tree
{
	int n,c[maxn];

	void init(int n)
	{
		this->n = n;
		memset(c,0,sizeof(c));
	}
	int lowbit(int x)
	{
		return x & -x;
	}
	void update(int x,int d)
	{
		while(x <= n)
		{
			c[x] += d;
			x += lowbit(x);
		}
	}
	int sum(int x)
	{
		int ans = 0;
		while(x > 0)
		{
			ans += c[x];
			x -= lowbit(x);
		}
		return ans;
	}
}tree;
int n,m,fa[maxn],cnt[maxn];

int find(int x)
{
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}

int main()
{
	int op;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		tree.init(n);
		tree.update(1,n);
		for(int i = 1; i <= n; i++) fa[i] = i,cnt[i] = 1;
		while(m--)
		{
			scanf("%d",&op);
			if(op == 0)
			{
				int i,j;
				scanf("%d%d",&i,&j);
				int f1 = find(i);
				int f2 = find(j);
				if(f1 != f2)
				{
					tree.update(cnt[f1],-1);
					tree.update(cnt[f2],-1);
					fa[f2] = f1;
					cnt[f1] += cnt[f2];
					tree.update(cnt[f1],1);
				}
			}
			else
			{
				int k,l = 1,r = n,mid,ans;
				scanf("%d",&k);
				while(l <= r)
				{ 
					mid = (l + r) >> 1;
					int tmp = tree.sum(n) - tree.sum(mid-1);
					if(tmp >= k)
					{
						ans = mid;
						l = mid + 1;
					}
					else r = mid - 1;
				}
				printf("%d\n",ans);
			}
		}
	}
	return 0;
}