poj2761（treap入门）

给n个数，然后m个询问，询问任意区间的第k小的数，特别的，任意两个区间不存在包含关系，

也就是说，将所有的询问按L排序之后， 对于i<j ,   Li < Lj 且 Ri < Rj

所以只要每次查询的时候，treap都是对应区间[Li,Ri] ，那么查找第k小就很简单了

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <time.h>
#include <algorithm>
using namespace std;
const int INF = 1<<30;
const int N = 100001 + 10;
struct Treap
{
    int ch[2];
    int fix,key;
    int same,cnt;
}treap[N*40];
struct Node
{
    int l, r,k,id;
    bool operator<(const Node&rhs)const
    {
        return l < rhs.l;
    }
}q[N*5];
int tot;
int val[N];
int ans[N*5];
void maintain(int o)
{
    int lCnt = treap[o].ch[0] == 0 ? 0 : treap[treap[o].ch[0]].cnt;
    int rCnt = treap[o].ch[1] == 0 ? 0 : treap[treap[o].ch[1]].cnt;
    treap[o].cnt = treap[o].same + lCnt + rCnt;
}
void rotate(int &o, int d)
{
    int t = treap[o].ch[d^1];
    treap[o].ch[d^1] = treap[t].ch[d];
    treap[t].ch[d] = o;
    maintain(o);
    maintain(t);
    o = t;
}
void insert(int &o, int tkey)
{
    if(o==0)
    {
        o = ++tot;
        if(tot>=N*60) while(true);;
        treap[o].ch[0] = treap[o].ch[1] = 0;
        treap[o].cnt = treap[o].same = 1;
        treap[o].key = tkey;
        treap[o].fix = rand();
        maintain(o);
    }
    else if(tkey < treap[o].key)
    {
        insert(treap[o].ch[0],tkey);
        maintain(o);
        if(treap[treap[o].ch[0]].fix > treap[o].fix)
            rotate(o,1);
    }
    else if(tkey > treap[o].key)
    {
        insert(treap[o].ch[1],tkey);
        maintain(o);
        if(treap[treap[o].ch[1]].fix > treap[o].fix)
            rotate(o,0);
    }
    else
    {
        treap[o].same++;
        maintain(o);
    }

}
void remove(int &o, int tkey)
{
    if(tkey < treap[o].key)
    {
        remove(treap[o].ch[0],tkey);
    }
    else if(tkey > treap[o].key)
    {
        remove(treap[o].ch[1],tkey);
    }
    else
    {
        if(treap[o].same!=1)
            treap[o].same--;
        else if(treap[o].ch[0]==0 && treap[o].ch[1]==0)
            o = 0;
        else if(treap[o].ch[0]==0)
        {
            rotate(o,0);
            remove(treap[o].ch[0],tkey);
        }
        else if(treap[o].ch[1]==0)
        {
            rotate(o,1);
            remove(treap[o].ch[1],tkey);
        }
        else if(treap[treap[o].ch[0]].fix <= treap[treap[o].ch[1]].fix)
        {
            rotate(o,0);
            remove(treap[o].ch[0],tkey);
        }
        else
        {
            rotate(o,1);
            remove(treap[o].ch[1],tkey);
        }
    }
    maintain(o);
}

int query(int o, int k)
{
    while(o)
    {
        int cnt = treap[o].ch[0]==0 ? 0 : treap[treap[o].ch[0]].cnt;
        if(cnt>=k)
            o = treap[o].ch[0];
        else if(cnt<k && k<=cnt+treap[o].same)
            return treap[o].key;
        else
        {

            k -= cnt + treap[o].same;
            o = treap[o].ch[1];
        }
    }
}
int main()
{
    //freopen("d:/in.txt","r",stdin);
    int n,m;
    srand(time(0));
    while(scanf("%d%d",&n,&m)!=EOF)
    {

        for(int i=1;i<=n;++i)
        {
            scanf("%d",&val[i]);
        }
        tot = 0;
        for(int i=0;i<m;++i)
        {
             scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].k);
             q[i].id = i;
        }
        sort(q,q+m);
        int root = 0;
        int L=1,R=1;//初始在Treap中的节点所属区间[L,R),注意半闭半开区间
        for(int i=0;i<m;i++)
        {
            while(L<q[i].l)
            {
                if(L<R) remove(root,val[L]);
                L++;
            }
            if(R<L) R=L;
            while(R<=q[i].r)
            {
                insert(root,val[R]);
                R++;
            }
            ans[q[i].id]=query(root,q[i].k);
        }
        for(int i=0;i<m;i++) printf("%d
",ans[i]);
    }
    return 0;
}