题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6534
Chika and Friendly Pairs
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 160 Accepted Submission(s): 52
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
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题意：找区间里a[i]-a[j]<=K的对数 i<j
思路：n和m都是27000 也就是2*10^4 1秒在10^7左右 所以我们还有10^3 普通莫队的复杂度是n^(3/2) 算一下27000当做25000的话 那么复杂度大概就是10^6 还有10可以用 显然用树结构来
维护求区间对数 树状数组复杂度log(n) 所以差不多就是10^7左右 果然是邀请赛的题，时间卡的是真的紧，，一点不让你浪费。 如果我们每次都用lower_bound和upperbound去找上下界
超时。。。 需要预处理上下界 这样就可以A这道题了
看代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
const LL INF=1e9+7;
const int maxn=27000+50;
LL N,M,K;
LL a[maxn],b[maxn];
LL c[maxn];
LL ans[maxn];
LL up[maxn],low[maxn];
LL block;
LL len;
struct query
{
LL l,r,num;
}q[maxn];
bool cmp(const query x,const query y)
{
if(x.l/block==y.l/block) return x.r<y.r;
return x.l<y.l;
}
int lowbit(int x)
{
return x&(-x);
}
int sea(LL x)//寻找大小为x的数在b数组中的下标
{
int l=1,r=len;
while(l<=r)
{
int mid=(l+r)>>1;
if(b[mid]==x) return mid;
else if(b[mid]>x)
{
r=mid-1;
}
else l=mid+1;
}
}
void update(LL x,int add)
{
while(x<=27000)
{
c[x]+=add;
x+=lowbit(x);
}
}
LL Query(LL x)
{
LL ret=0;
while(x>0)
{
ret+=c[x];
x-=lowbit(x);
}
return ret;
}
int main()
{
scanf("%lld%lld%lld",&N,&M,&K);
block=sqrt(N);
for(int i=1;i<=N;i++)
{
scanf("%lld",&a[i]);
b[i]=a[i];
}
// sort(a+1,a+1+N,cmp);
sort(b+1,b+N+1);
len=unique(b+1,b+N+1)-(b+1);//去重
for(int i=1;i<=M;i++)
{
scanf("%lld%lld",&q[i].l,&q[i].r);
q[i].num=i;
}
sort(q+1,q+M+1,cmp);
for(int i=1;i<=N;i++)
{
LL x=lower_bound(b+1,b+1+len,a[i]-K)-b;
LL y=upper_bound(b+1,b+1+len,a[i]+K)-b-1;
up[i]=y;
low[i]=x-1;
}
LL L=1,R=0;
LL ret=0;
for(int i=1;i<=M;i++)
{
while(R<q[i].r)//区间增大
{
// LL x=lower_bound(b+1,b+1+len,a[R+1]-K)-b;
// LL y=upper_bound(b+1,b+1+len,a[R+1]+K)-b-1;
//现在在b数组[x,y]之间的所有的数都是符合条件的
ret+=Query(up[R+1])-Query(low[R+1]);
update(sea(a[R+1]),1);//这个必须放在上面的下面 因为如果先放的话本身的值岂不是算进去了
R++;
}
while(R>q[i].r)//区间减小
{
update(sea(a[R]),-1);//这个必须放在下面的上面 因为你要减去它的影响 自然也不能算本身-本身
// LL x=lower_bound(b+1,b+1+len,a[R]-K)-b;
// LL y=upper_bound(b+1,b+1+len,a[R]+K)-b-1;
ret-=Query(up[R])-Query(low[R]);
R--;
}
while(L<q[i].l)//区间减小
{
update(sea(a[L]),-1);
// solve(L,-1);
// LL x=lower_bound(b+1,b+1+len,a[L]-K)-b;
// LL y=upper_bound(b+1,b+1+len,a[L]+K)-b-1;
ret-=Query(up[L])-Query(low[L]);
L++;
}
while(L>q[i].l)//区间增大
{
// LL x=lower_bound(b+1,b+1+len,a[L-1]-K)-b;
// LL y=upper_bound(b+1,b+1+len,a[L-1]+K)-b-1;
ret+=Query(up[L-1])-Query(low[L-1]);
update(sea(a[L-1]),1);
L--;
}
ans[q[i].num]=ret;
}
for(int i=1;i<=M;i++) printf("%lld\n",ans[i]);
//cout<<ans[i]<<endl;
return 0;
}
原文地址：https://www.cnblogs.com/caijiaming/p/10896352.html