HDU 5178 pairs（二分 STL lower_bound）

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本文介绍了一个计算二维平面上特定条件下点对数量的问题。输入一系列位于X轴上的点坐标，通过算法找出所有符合条件(|x[b]−x[a]|≤k, a<b)的点对数目。该算法使用了排序和二分查找技术来提高效率。

pairs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2038 Accepted Submission(s): 733

Problem Description

John has

n points on the X axis, and their coordinates are

(x[i],0),(i=0,1,2,…,n−1). He wants to know how many pairs

<a,b> that

|x[b]−x[a]|≤k.(a<b)

Input

The first line contains a single integer

T (about 5), indicating the number of cases.
Each test case begins with two integers

n,k(1≤n≤100000,1≤k≤109).
Next

n lines contain an integer

x[i](−109≤x[i]≤109), means the X coordinates.

Output

For each case, output an integer means how many pairs

<a,b> that

|x[b]−x[a]|≤k.

Sample Input

Sample Output

3
10

#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1000];
int cmp(int a,int b)
{
	return  a<b;
}
int main(int argc, char *argv[])
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,k;
		scanf("%d %d",&n,&k);
		int i;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}
		sort(a+1,a+n+1,cmp);
	    long long sum=0;
		for(i=1;i<=n;i++)
		{
			int l=lower_bound(a+i+1,a+n+1,a[i]-k)-a-1;
			int r=lower_bound(a+i+1,a+n+1,a[i]+k)-a-1;
			sum+=r-l;
		} 	 	
		printf("%lld\n",sum);
	}	
	return 0;
}