HDU 5178 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 )

 

 

Input

The first line contains a single integer ), means the X coordinates.

 

 

Output

For each case, output an integer means how many pairsk.

 

 

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

 

首先将所给的坐标从大到小排序，则此题转化为：对排序后的新数列，对每个左边的x[a]找到它右边最远的x[b]使得x[b]-x[a]<=k，累计所有的b-a的和即可。

这里要做一个小优化，否则直接暴搜会TLE：

　　用一个数组pos[i]记录x[i]右边最远的x[b]的b值，由于x[b]是单调递增的，所以找pos[i+1]时只要将b从pos[i]继续往右找即可。

另外答案结果非常大，应用long long来保存

 

 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++]