HDU 5178 pairs —— 思维 + 二分

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5178

 

pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3090    Accepted Submission(s): 1085

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

2 5 5 -100 0 100 101 102 5 300 -100 0 100 101 102

 

 

Sample Output

3 10

 

 

题解：

两个要求：1:b>a   2.abs(x[b]-x[a])<=k。

先对数组进行排序。对于满足上述条件的a、b（ab为排序前的序号）来说。a、b的关系是可以互相转化的：

当abs(x[b]-x[a])<=k 且 b>a时，那这一对a、b固然满足条件。

当abs(x[b]-x[a])<=k 且 a>b时，那a就变成b，b就变成a，所以也满足条件。

综上，只需要找到abs(x[b]-x[a])<=k 的a、b对即可，无所谓ab的大小关系。但是直接这样计算会出现重复，正确的做法是，只取一边，即：x[a]+k或者是 x[a]-k，这样就能避免重复。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define rep(i,s,t) for(int (i)=(s); (i)<=(t); (i)++)
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const double eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 1e5+10;

int n, k;
int a[maxn];

int sch(int x)
{
    int l = 1, r = n;
    while(l<=r)
    {
        int mid = (l+r)>>1;
        if(a[mid]<=x)
            l = mid + 1;
        else
            r = mid - 1;
    }
    return r;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        rep(i,1,n)
            scanf("%d",&a[i]);

        sort(a+1,a+1+n);
        LL ans = 0;
        rep(i,1,n-1)
        {
            int p = sch(a[i]+k);
//            int p = upper_bound(a+1,a+1+n,a[i]+k) - (a+1);
            ans += p-i;
        }
        printf("%I64d\n",ans);
    }
}