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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: 

Description

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;
}






SPOJ 3267. D-query (主席树or树状数组离线)