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

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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\n",query(T[l],r));

117         }

118     }

119     return 0;

120 }