Chika and Friendly Pairs（莫队+离散化+树状数组）

Chika and Friendly Pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1164    Accepted Submission(s): 421

Problem Description

Chika gives you an integer sequence a1,a2,…,an and m tasks. For each task, you need to answer the number of "friendly pairs" in a given interval.

friendly pair: for two integers ai and aj, if i<j and the absolute value of ai−aj is no more than a given constant integer K, then (i,j) is called a "friendly pair".A friendly pair (i,j) in a interval [L,R] should satisfy L≤i<j≤R.

 

Input

The first line contains 3 integers n (1≤n≤27000), m (1≤m≤27000) and K (1≤K≤109), representing the number of integers in the sequence a, the number of tasks and the given constant integer.
The second line contains n non-negative integers, representing the integers in the sequence a. Every integer of sequence a is no more than 109.
Then m lines follow, each of which contains two integers L, R (1≤L≤R≤n). The meaning is to ask the number of "friendly pairs" in the interval [L,R]。

 

Output

For each task, you need to print one line, including only one integer, representing the number of "friendly pairs" in the query interval.

 

Sample Input

7 5 3 2 5 7 5 1 5 6 6 6 1 3 4 6 2 4 3 4

 

Sample Output

0 2 1 3 1

 

Source

2019CCPC湖南全国邀请赛（广东省赛、江苏省赛）重现赛

 

解题思路：莫队维护区间，树状数组维护小于等于这个数的个数,  由于数很大要离散化.

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

typedef long long ll;

int n,m,k,L,R;
const int N=27007;
int arr[N],brr[N],low[N],up[N],my[N],cnt[N];
int ee[N],res;


inline int read(){
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    return x*f;
}

int lowbits(int x){
    return x&(-x);
}

void add(int x,int num){
    while(x<27000){
        cnt[x]+=num;
        x+=lowbits(x);
    }
}

int sum(int x){
    int res=0;
    while(x){
        res+=cnt[x];
        x-=lowbits(x);
    }
    return res;
}

struct Node{
    int l,r,id,ans;
    bool operator<(const Node&X)const{
//        if(ans!=X.ans) return ans<X.ans;
//        return r<X.r;
        return (ans^X.ans)?l<X.l:(ans&1)?r<X.r:r>X.r;
    }
}A[N];


int main(){
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++) scanf("%d",&arr[i]),brr[i]=arr[i];
    sort(arr+1,arr+1+n);
    int size=unique(arr+1,arr+1+n)-arr-1;
    for(int i=1;i<=n;i++){
        low[i]=lower_bound(arr+1,arr+1+size,brr[i]-k)-arr-1;  //比x小的
        up[i]=upper_bound(arr+1,arr+1+size,brr[i]+k)-arr-1;  //等于y的
        my[i]=lower_bound(arr+1,arr+1+size,brr[i])-arr;
    }
    int big=sqrt(n);
    for(int i=1;i<=m;i++){
        A[i].l=read(),A[i].r=read();
        A[i].id=i;A[i].ans=(A[i].l-1)/big+1;    //分块在哪个块
    }
    sort(A+1,A+1+m);
    L=1,R=0;
    for(int i=1;i<=m;i++){
        while(L<A[i].l){
            add(my[L],-1);
            res-=sum(up[L])-sum(low[L]);
            L++;
        }
        while(L>A[i].l){
            L--;
            res+=sum(up[L])-sum(low[L]);
            add(my[L],1);
        }
        while(R>A[i].r){
            add(my[R],-1);
            res-=sum(up[R])-sum(low[R]);
            R--;
        }
        while(R<A[i].r){
            R++;
            res+=sum(up[R])-sum(low[R]);
            add(my[R],1);
        }
        ee[A[i].id]=res;
    }
    for(int i=1;i<=m;i++) printf("%d
",ee[i]);
    return 0;
}