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++]

 

posted @ 2015-02-28 22:08  ~~Snail~~  阅读(742)  评论(1)  编辑  收藏  举报