SPOJ 3267. D-query (主席树or树状数组离线）

A - D-query
Time Limit:1500MS     Memory Limit:0KB     64bit IO Format:%lld
& %llu
Submit Status Practice SPOJ
DQUERY

Appoint description: 
System Crawler  (2014-12-06)

Description

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

就是询问区间里不同元素的个数，可以用树状数组离线操作或者用主席树

主席树版：

/*************************************************************************
> File Name: cf.cpp
> Author: acvcla
> QQ:
> Mail: acvcla@gmail.com
> Created Time: 1949Äê10ÔÂ1ÈÕ ÐÇÆÚÒ» 0ʱ0·Ö0Ãë
************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 3e4 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int tot,n,m,date[maxn],root[maxn];
struct chairTnode
{
int s,rc,lc;
chairTnode(int s=0,int lc=0,int rc=0):s(s),lc(lc),rc(rc){}
};
chairTnode chairT[maxn*20];
int newnode(int sum,int lson,int rson)
{
int rt=++tot;
chairT[rt]=chairTnode(sum,lson,rson);
return rt;
}
void Insert(int &rt,int pre_rt,int pos,int L,int R,int val)
{
chairTnode &t=chairT[pre_rt];
rt=newnode(t.s+val,t.lc,t.rc);
if(L==R)return;
int mid=(L+R)>>1;
if(pos<=mid)Insert(chairT[rt].lc,t.lc,pos,L,mid,val);
else Insert(chairT[rt].rc,t.rc,pos,mid+1,R,val);
}
int Query(int rt,int L,int R,int pos)//Query sum of [pos,R]
{
if(L==pos)return chairT[rt].s;
int mid=(L+R)>>1;
if(pos<=mid)return Query(chairT[rt].lc,L,mid,pos)+chairT[chairT[rt].rc].s;
return Query(chairT[rt].rc,mid+1,R,pos);
}
int main()
{
while(~scanf("%d",&n))
{
rep(i,1,n){
scanf("%d",date+i);
}
tot=root[0]=0;
map<int,int>q;
int t;
for(int i=1;i<=n;i++){
if(!q[date[i]]){
Insert(root[i],root[i-1],i,1,n,1);//在位置i的数目加1
}else{
Insert(t,root[i-1],q[date[i]],1,n,-1);//之前已经出现过，减去前面加的那次
Insert(root[i],t,i,1,n,1);//加到现在这次来
}
q[date[i]]=i;
}
int ql,qr;
scanf("%d",&m);
while(m--)
{
scanf("%d%d",&ql,&qr);
printf("%d\n",Query(root[qr],1,n,ql));
}
}
return 0;
}

树状数组版：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<map>
const int maxn1=3e4+5;
const int maxn2=2e5+5;
typedef long long LL;
using namespace std;
int l[maxn2],r[maxn2],loc[maxn1],A[maxn1],id[maxn2],n;
int C[maxn1],ans[maxn2];
inline int lowbit(int x){return x&-x;}
void update(int x,int d)
{
while(x<=n)
{
C[x]+=d;
x+=lowbit(x);
}
}
int Query(int x)
{
int sum=0;
while(x>0)
{
sum+=C[x];
x-=lowbit(x);
}
return sum;
}
bool cmp(int i,int j)
{
return r[i]<r[j];
}
int  main()
{
while(~scanf("%d",&n))
{
memset(loc,0,sizeof(loc));
memset(C,0,sizeof(C));
map<int,int>q;
q.clear();
for(int i=1;i<=n;i++)
{
scanf("%d",A+i);
update(i,1);
if(!q[A[i]])
{
q[A[i]]=i;
}
}
int m;
scanf("%d",&m);
for(int i=1;i<=m;i++)
{
scanf("%d%d",l+i,r+i);
id[i]=i;
}
sort(id+1,id+1+m,cmp);
/*主要思想是将计数尽可能拖到后面来*/
int Right=1;
for (int i = 1; i <= m; ++i)
{
int x=id[i];
while(Right<=r[x])
{
if(q[A[Right]]!=Right){
update(q[A[Right]],-1);/*在后面出现了,让前面的计数抵消，在后面计数，保证查询区间里的数不遗漏*/
q[A[Right]]=Right;
}
++Right;
}
Right=r[x];
ans[x]=Query(r[x])-Query(l[x]-1);
}
for(int i=1;i<=m;i++)printf("%d\n",ans[i]);
}
return 0;
}