SPOJ 3267 D-query （可持久化线段树，区间重复元素个数）

D-query

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 

题意：求区间内不重复的数的个数。 n,m<=100000
题解：建立可持久化线段树，以右端点为最后建立现在版本线段树，
　　　然后就是维护每一棵线段树，就是前面的点什么时候失效，询问大区间就一定会包含小区间中的
　　　相同权值的点，然后只需要记录和即可，离散化还是需要的+二分。

#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstdio>

#define N 60007
#define M 20000007
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if (ch=='-') f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}

int n,top,sz,q;
int a[N],b[N],num[N],pos[N],root[N];
int ls[M],rs[M],sum[M];

int bs(int num)
{
    int l=1,r=top;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if (b[mid]==num) return mid;
        if (b[mid]<num) l=mid+1;
        else r=mid-1;
    }
}
void change(int l,int r,int x,int &y,int wei,int z)
{
    y=++sz;
    if (l==r)
    {
        sum[y]=sum[x]+z;
        return;
    }
    ls[y]=ls[x],rs[y]=rs[x],sum[y]=sum[x]+z;
    int mid=(l+r)>>1;
    if (wei<=mid) change(l,mid,ls[x],ls[y],wei,z);
    else change(mid+1,r,rs[x],rs[y],wei,z);
}

int query(int p,int l,int r,int x,int y)
{
    if (l==x&&y==r) return sum[p];
    int mid=(l+r)>>1;
    if (y<=mid) return query(ls[p],l,mid,x,y);
    else if (x>mid) return query(rs[p],mid+1,r,x,y);
    else return query(ls[p],l,mid,x,mid)+query(rs[p],mid+1,r,mid+1,y);
}
int main()
{
    int n=read();
    for (int i=1;i<=n;i++)
        a[i]=read(),b[i]=a[i];
    sort(b+1,b+n+1);
    top=1;
    for (int i=2;i<=n;i++)
        if (b[i]!=b[i-1]) b[++top]=b[i];
    for (int i=1;i<=n;i++)
    {
        int num=bs(a[i]);
        change(1,n,root[i-1],root[i],i,1);
        if (pos[num]) change(1,n,root[i],root[i],pos[num],-1);
        pos[num]=i;
    }
    q=read();
    while(q--)
    {
        int l=read(),r=read();
        printf("%d
",query(root[r],1,n,l,r));
    }
}