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
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.
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<cmath>
6 using namespace std;
7
8 typedef long long ll;
9
10 int n,m,k,L,R;
11 const int N=27007;
12 int arr[N],brr[N],low[N],up[N],my[N],cnt[N];
13 int ee[N],res;
14
15
16 inline int read(){
17 int x=0,f=1;
18 char ch=getchar();
19 while(ch<‘0‘||ch>‘9‘){
20 if(ch==‘-‘)
21 f=-1;
22 ch=getchar();
23 }
24 while(ch>=‘0‘&&ch<=‘9‘){
25 x=(x<<1)+(x<<3)+(ch^48);
26 ch=getchar();
27 }
28 return x*f;
29 }
30
31 int lowbits(int x){
32 return x&(-x);
33 }
34
35 void add(int x,int num){
36 while(x<27000){
37 cnt[x]+=num;
38 x+=lowbits(x);
39 }
40 }
41
42 int sum(int x){
43 int res=0;
44 while(x){
45 res+=cnt[x];
46 x-=lowbits(x);
47 }
48 return res;
49 }
50
51 struct Node{
52 int l,r,id,ans;
53 bool operator<(const Node&X)const{
54 // if(ans!=X.ans) return ans<X.ans;
55 // return r<X.r;
56 return (ans^X.ans)?l<X.l:(ans&1)?r<X.r:r>X.r;
57 }
58 }A[N];
59
60
61 int main(){
62 scanf("%d%d%d",&n,&m,&k);
63 for(int i=1;i<=n;i++) scanf("%d",&arr[i]),brr[i]=arr[i];
64 sort(arr+1,arr+1+n);
65 int size=unique(arr+1,arr+1+n)-arr-1;
66 for(int i=1;i<=n;i++){
67 low[i]=lower_bound(arr+1,arr+1+size,brr[i]-k)-arr-1; //比x小的
68 up[i]=upper_bound(arr+1,arr+1+size,brr[i]+k)-arr-1; //等于y的
69 my[i]=lower_bound(arr+1,arr+1+size,brr[i])-arr;
70 }
71 int big=sqrt(n);
72 for(int i=1;i<=m;i++){
73 A[i].l=read(),A[i].r=read();
74 A[i].id=i;A[i].ans=(A[i].l-1)/big+1; //分块在哪个块
75 }
76 sort(A+1,A+1+m);
77 L=1,R=0;
78 for(int i=1;i<=m;i++){
79 while(L<A[i].l){
80 add(my[L],-1);
81 res-=sum(up[L])-sum(low[L]);
82 L++;
83 }
84 while(L>A[i].l){
85 L--;
86 res+=sum(up[L])-sum(low[L]);
87 add(my[L],1);
88 }
89 while(R>A[i].r){
90 add(my[R],-1);
91 res-=sum(up[R])-sum(low[R]);
92 R--;
93 }
94 while(R<A[i].r){
95 R++;
96 res+=sum(up[R])-sum(low[R]);
97 add(my[R],1);
98 }
99 ee[A[i].id]=res;
100 }
101 for(int i=1;i<=m;i++) printf("%d\n",ee[i]);
102 return 0;
103 }
原文地址：https://www.cnblogs.com/qq-1585047819/p/11369271.html