题目描述
English VietnameseGiven a sequence of n numbers a _{1}1? , a _{2}2? , ..., a _{n}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}i? , a _{i+1}i+1? , ..., a _{j}j?.
输入输出格式
输入格式：
- Line 1: n (1 ≤ n ≤ 30000).
- Line 2: n numbers a _{1}1? , a _{2}2? , ..., a _{n}n? (1 ≤ a _{i}i? ≤ 10 ^{6}6 ).
- 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).
输出格式：
- For each d-query (i, j), print the number of distinct elements in the subsequence a _{i}i? , a _{i+1}i+1? , ..., a _{j}j? in a single line.
输入输出样例
输出样例#1：
3 2 3
离线扫描经典题，但是我复习一下主席树233
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=200005;
struct node{
int tag;
node *lc,*rc;
}nil[maxn*53],*rot[30005],*cnt;
int n,Q,a[30005],le,ri,pre[maxn*5];
node *update(node *u,int l,int r){
node *ret=++cnt;
*ret=*u;
if(l>=le&&r<=ri){
ret->tag++;
return ret;
}
int mid=l+r>>1;
if(le<=mid) ret->lc=update(ret->lc,l,mid);
if(ri>mid) ret->rc=update(ret->rc,mid+1,r);
return ret;
}
int query(node *u,int l,int r){
if(l==r) return u->tag;
int mid=l+r>>1;
if(le<=mid) return u->tag+query(u->lc,l,mid);
else return u->tag+query(u->rc,mid+1,r);
}
inline void init(){
rot[0]=nil->lc=nil->rc=cnt=nil;
nil->tag=0;
for(int i=1;i<=n;i++){
le=pre[a[i]]+1,ri=i,rot[i]=update(rot[i-1],1,1000000);
pre[a[i]]=i;
}
}
inline void solve(){
scanf("%d",&Q);
while(Q--){
scanf("%d%d",&le,&ri);
printf("%d\n",query(rot[ri],1,1000000));
}
}
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",a+i);
init();
solve();
return 0;
}
