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