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

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

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-9-5 23:54:37
  4 File Name     :F:2013ACM练习专题学习主席树SPOJ_DQUERY.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 
 21 /*
 22  * 给出一个序列，查询区间内有多少个不相同的数
 23  */
 24 const int MAXN = 30010;
 25 const int M = MAXN * 100;
 26 int n,q,tot;
 27 int a[MAXN];
 28 int T[M],lson[M],rson[M],c[M];
 29 int build(int l,int r)
 30 {
 31     int root = tot++;
 32     c[root] = 0;
 33     if(l != r)
 34     {
 35         int mid = (l+r)>>1;
 36         lson[root] = build(l,mid);
 37         rson[root] = build(mid+1,r);
 38     }
 39     return root;
 40 }
 41 int update(int root,int pos,int val)
 42 {
 43     int newroot = tot++, tmp = newroot;
 44     c[newroot] = c[root] + val;
 45     int l = 1, r = n;
 46     while(l < r)
 47     {
 48         int mid = (l+r)>>1;
 49         if(pos <= mid)
 50         {
 51             lson[newroot] = tot++; rson[newroot] = rson[root];
 52             newroot = lson[newroot]; root = lson[root];
 53             r = mid;
 54         }
 55         else
 56         {
 57             rson[newroot] = tot++; lson[newroot] = lson[root];
 58             newroot = rson[newroot]; root = rson[root];
 59             l = mid+1;
 60         }
 61         c[newroot] = c[root] + val;
 62     }
 63     return tmp;
 64 }
 65 int query(int root,int pos)
 66 {
 67     int ret = 0;
 68     int l = 1, r = n;
 69     while(pos < r)
 70     {
 71         int mid = (l+r)>>1;
 72         if(pos <= mid)
 73         {
 74             r = mid;
 75             root = lson[root];
 76         }
 77         else
 78         {
 79             ret += c[lson[root]];
 80             root = rson[root];
 81             l = mid+1;
 82         }
 83     }
 84     return ret + c[root];
 85 }
 86 
 87 int main()
 88 {
 89     //freopen("in.txt","r",stdin);
 90     //freopen("out.txt","w",stdout);
 91     while(scanf("%d",&n) == 1)
 92     {
 93         tot = 0;
 94         for(int i = 1;i <= n;i++)
 95             scanf("%d",&a[i]);
 96         T[n+1] = build(1,n);
 97         map<int,int>mp;
 98         for(int i = n;i>= 1;i--)
 99         {
100             if(mp.find(a[i]) == mp.end())
101             {
102                 T[i] = update(T[i+1],i,1);
103             }
104             else
105             {
106                 int tmp = update(T[i+1],mp[a[i]],-1);
107                 T[i] = update(tmp,i,1);
108             }
109             mp[a[i]] = i;
110         }
111         scanf("%d",&q);
112         while(q--)
113         {
114             int l,r;
115             scanf("%d%d",&l,&r);
116             printf("%d
",query(T[l],r));
117         }
118     }
119     return 0;
120 }