SPOJ DQUERY:D-query

主席树/树状数组。给一个区间，多次询问[l,r]内有多少个不重复的元素。每个前缀都建线段树，询问直接r的[l,r]就可以了。（似乎对主席树有一点了解了？。。。话说spoj好高级的样子。。。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define REP(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=30005;
const int inf=0x7f7f7f7f;
struct node{
	node *l,*r;int sum;
};
node nodes[nmax*25],*pt=nodes,*root[nmax];
int a[nmax],b[nmax],vis[nmax*40];
node* build(int l,int r){
	node* op=pt++;op->sum=0;
	if(l==r) return op;
	int mid=(l+r)>>1;
	op->l=build(l,mid);op->r=build(mid+1,r);
	return op;
}
node* update(int p,int add,node* t,int l,int r){
	node* op=pt++;op->sum=t->sum+add;
	if(l==r) return op;
	int mid=(l+r)>>1;
	if(p<=mid) op->r=t->r,op->l=update(p,add,t->l,l,mid);
	else op->l=t->l,op->r=update(p,add,t->r,mid+1,r);
	return op;
}
int query(int tl,int tr,int l,int r,node* t){
	if(tl<=l&&tr>=r) return t->sum;
	int ans=0,mid=(l+r)>>1;
	if(tl<=mid) ans+=query(tl,tr,l,mid,t->l);
	if(tr>mid) ans+=query(tl,tr,mid+1,r,t->r);
	return ans;
}
int main(){
	int n,m;
	while(scanf("%d",&n)!=EOF){
		pt=nodes;clr(vis,0);
		REP(i,1,n) a[i]=read();

		root[0]=build(1,n);
		REP(i,1,n){
			node* t=update(i,1,root[i-1],1,n);
			if(vis[a[i]]) root[i]=update(vis[a[i]],-1,t,1,n);
			else root[i]=t;
			vis[a[i]]=i;
		}

		m=read();
		REP(i,1,m){
			int s=read(),t=read();
			printf("%d\n",query(s,t,1,n,root[t]));
		}
	}
	return 0;
}

　　

D-query

Time Limit: 227MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

Submit Status

Description

English

Vietnamese

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

Hint

Added by:	Duc
Date:	2008-10-26
Time limit:	0.227s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel G860)
Languages:	All except: ERL JS NODEJS PERL 6 VB.net
Resource:	© VNOI