HDU 5178 Pairs

pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 96    Accepted Submission(s): 38

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

 

 

Source

BestCoder Round #31

 

首先將所給的座標從大到小排序，則此題轉化爲：對排序後的新數列，對每一個左邊的x[a]找到它右邊最遠的x[b]使得x[b]-x[a]<=k，累計全部的b-a的和便可。

這裏要作一個小優化，不然直接暴搜會TLE：

　　用一個數組pos[i]記錄x[i]右邊最遠的x[b]的b值，因爲x[b]是單調遞增的，因此找pos[i+1]時只要將b從pos[i]繼續往右找便可。

另外答案結果很是大，應用long long來保存

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int kase;
10     scanf ("%d", &kase);
11     while (kase--)
12     {
13         int n, k;
14         long long ans = 0;
15         int num[100200], pos[100200];
16         scanf ("%d %d", &n, &k);
17         for (int i = 0; i < n; i++)
18         {
19             scanf ("%d", &num[i]);
20         }
21         sort (num, num + n);
22         for (int i = 0; i < n; i++)
23         {
24             if (i == 0)
25             {
26                 for (int j = i + 1; j <= n; j++)
27                     if ( (j < n && num[j] - num[i] > k) || (j == n))
28                     {
29                         pos[i] = j - 1;
30                         ans += j - i - 1;
31                         break;
32                     }
33             }
34             else
35             {
36                 for (int j = pos[i - 1]; j <= n; j++)
37                     if ( (j < n && num[j] - num[i] > k) || j == n)
38                     {
39                         pos[i] = j - 1;
40                         ans += j - i - 1;
41                         break;
42                     }
43             }
44         }
45         printf ("%I64d\n", ans);
46     }
47     return 0;
48 }

[C++]