SPOJ 3267 DQUERY - D-query （主席树）（区间数的种数）

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
,tot;
vector<int>pos;
int arr
;
int last_pos
;

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\n",query(root[l-1],root[r],1,n,l,r));
}
}
return 0;
}