SPOJ 3267. D-query （主席树，查询区间有多少个不相同的数）

3267. D-query Problem code: DQUERY

English

Vietnamese

Given a sequence of n numbers a₁, a₂, ..., a_n and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the subsequence a_i, a_i+1, ..., a_j.

Input

• Line 1: n (1 ≤ n ≤ 30000).
• Line 2: n numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁶).
• Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.
• In the next q lines, each line contains 2 numbers i, j representing a d-query (1 ≤ i ≤ j ≤ n).

Output

• For each d-query (i, j), print the number of distinct elements in the subsequence a_i, a_i+1, ..., a_j in a single line.

Example

Input
5
1 1 2 1 3
3
1 5
2 4
3 5

Output
3
2
3

主席树的入门题了

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013-9-5 23:54:37
 File Name     :F:\2013ACM练习\专题学习\主席树\SPOJ_DQUERY.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 /*
  * 给出一个序列，查询区间内有多少个不相同的数
  */
 const int MAXN = ;
 const int M = MAXN * ;
 int n,q,tot;
 int a[MAXN];
 int T[M],lson[M],rson[M],c[M];
 int build(int l,int r)
 {
     int root = tot++;
     c[root] = ;
     if(l != r)
     {
         int mid = (l+r)>>;
         lson[root] = build(l,mid);
         rson[root] = build(mid+,r);
     }
     return root;
 }
 int update(int root,int pos,int val)
 {
     int newroot = tot++, tmp = newroot;
     c[newroot] = c[root] + val;
     int l = , r = n;
     while(l < r)
     {
         int mid = (l+r)>>;
         if(pos <= mid)
         {
             lson[newroot] = tot++; rson[newroot] = rson[root];
             newroot = lson[newroot]; root = lson[root];
             r = mid;
         }
         else
         {
             rson[newroot] = tot++; lson[newroot] = lson[root];
             newroot = rson[newroot]; root = rson[root];
             l = mid+;
         }
         c[newroot] = c[root] + val;
     }
     return tmp;
 }
 int query(int root,int pos)
 {
     int ret = ;
     int l = , r = n;
     while(pos < r)
     {
         int mid = (l+r)>>;
         if(pos <= mid)
         {
             r = mid;
             root = lson[root];
         }
         else
         {
             ret += c[lson[root]];
             root = rson[root];
             l = mid+;
         }
     }
     return ret + c[root];
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     while(scanf("%d",&n) == )
     {
         tot = ;
         for(int i = ;i <= n;i++)
             scanf("%d",&a[i]);
         T[n+] = build(,n);
         map<int,int>mp;
         for(int i = n;i>= ;i--)
         {
             if(mp.find(a[i]) == mp.end())
             {
                 T[i] = update(T[i+],i,);
             }
             else
             {
                 int tmp = update(T[i+],mp[a[i]],-);
                 T[i] = update(tmp,i,);
             }
             mp[a[i]] = i;
         }
         scanf("%d",&q);
         while(q--)
         {
             int l,r;
             scanf("%d%d",&l,&r);
             printf("%d\n",query(T[l],r));
         }
     }
     return ;
 }