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 = <<;
 const int N =  + ;
 struct Treap
 {
     int ch[];
     int fix,key;
     int same,cnt;
 }treap[N*];
 struct Node
 {
     int l, r,k,id;
     bool operator<(const Node&rhs)const
     {
         return l < rhs.l;
     }
 }q[N*];
 int tot;
 int val[N];
 int ans[N*];
 void maintain(int o)
 {
     int lCnt = treap[o].ch[] ==  ?  : treap[treap[o].ch[]].cnt;
     int rCnt = treap[o].ch[] ==  ?  : treap[treap[o].ch[]].cnt;
     treap[o].cnt = treap[o].same + lCnt + rCnt;
 }
 void rotate(int &o, int d)
 {
     int t = treap[o].ch[d^];
     treap[o].ch[d^] = treap[t].ch[d];
     treap[t].ch[d] = o;
     maintain(o);
     maintain(t);
     o = t;
 }
 void insert(int &o, int tkey)
 {
     if(o==)
     {
         o = ++tot;
         if(tot>=N*) while(true);;
         treap[o].ch[] = treap[o].ch[] = ;
         treap[o].cnt = treap[o].same = ;
         treap[o].key = tkey;
         treap[o].fix = rand();
         maintain(o);
     }
     else if(tkey < treap[o].key)
     {
         insert(treap[o].ch[],tkey);
         maintain(o);
         if(treap[treap[o].ch[]].fix > treap[o].fix)
             rotate(o,);
     }
     else if(tkey > treap[o].key)
     {
         insert(treap[o].ch[],tkey);
         maintain(o);
         if(treap[treap[o].ch[]].fix > treap[o].fix)
             rotate(o,);
     }
     else
     {
         treap[o].same++;
         maintain(o);
     }

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

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

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

         for(int i=;i<=n;++i)
         {
             scanf("%d",&val[i]);
         }
         tot = ;
         for(int i=;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 = ;
         int L=,R=;//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=;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=;i<m;i++) printf("%d\n",ans[i]);
     }
     return ;
 }