SPOJ 3267 DQUERY

DQUERY - D-query

#sorting #tree

English

Vietnamese

Given a sequence of n numbers a1, a2, ..., an 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 ai, ai+1, ..., aj.

Input

• Line 1: n (1 ≤ n ≤ 30000).
• Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 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 ai, ai+1, ..., aj in a single line.

   

Example

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

Output
3
2
3 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define lson(x) ((x<<1))
#define rson(x) ((x<<1)+1)
using namespace std;
typedef long long ll;
const int N=5e4+50;
const int M=N*N+10;
struct seg {
    int lson,rson;
    int cnt;
};
seg T[N*20];
int root[N],tot;
vector<int>pos;
int arr[N];
int last_pos[N];

void init() {
    pos.clear();
    met(root,0);met(last_pos,0);
    tot=0;
    T[0].cnt=T[0].lson=T[0].rson=0;
}
void update(int &cur,int ori,int l,int r,int pos,int flag) {
    cur=++tot;
    T[cur]=T[ori];
    T[cur].cnt+=flag;
    if(l==r)
        return ;
    int mid=(l+r)/2;
    if(pos<=mid)
        update(T[cur].lson,T[ori].lson,l,mid,pos,flag);
    else
        update(T[cur].rson,T[ori].rson,mid+1,r,pos,flag);
}
int query(int S,int E,int l,int r,int x,int y) {
    if(x<=l&&r<=y)
        return T[E].cnt-T[S].cnt;
    else {
        int mid=(l+r)/2;
        if(y<=mid)
            return query(T[S].lson,T[E].lson,l,mid,x,y);
        else if(x>mid)
            return query(T[S].rson,T[E].rson,mid+1,r,x,y);
        else
            return query(T[S].lson,T[E].lson,l,mid,x,mid)+query(T[S].rson,T[E].rson,mid+1,r,mid+1,y);
    }
}
int main(void) {
    int n,m,i,l,r;
    while (~scanf("%d",&n)) {
        init();
        for (i=1; i<=n; ++i) {
            scanf("%d",&arr[i]);
            pos.push_back(arr[i]);
        }
        scanf("%d",&m);
        sort(pos.begin(),pos.end());
        pos.erase(unique(pos.begin(),pos.end()),pos.end());
        int temp_rt=0;
        for (i=1; i<=n; ++i) {
            arr[i]=lower_bound(pos.begin(),pos.end(),arr[i])-pos.begin()+1;
            if(!last_pos[arr[i]]) {
                update(root[i],root[i-1],1,n,i,1);
                last_pos[arr[i]]=i;
            } else {
                update(temp_rt,root[i-1],1,n,last_pos[arr[i]],-1);
                update(root[i],temp_rt,1,n,i,1);
                last_pos[arr[i]]=i;
            }
        }
        for (i=0; i<m; ++i) {
            scanf("%d%d",&l,&r);
            printf("%d
",query(root[l-1],root[r],1,n,l,r));
        }
    }
    return 0;
}