Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5716    Accepted Submission(s): 1884

Problem Description

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

 

 

Input

There are multiple test cases. For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range [1, 100000]. m and k are in the range [0, 1000000]. The second line has n integers, which are all in the range [0, 1000000]. Proceed to the end of file.

 

 

Output

For each test case, print the length of the subsequence on a single line.

 

 

Sample Input

5 0 0 1 1 1 1 1 5 0 3 1 2 3 4 5

 

 

Sample Output

5 4

 

题解：

一个序列的最大值与最小值的差在m和k之间；

单调队列。

维护最大值和最小值，如果发现最大值和最小值的差大于k，那么就移动下标最靠前的队列。

注意如下数据：

5 2 4

2 1 5 2 2

应该用一个last标记上一个移动的位置，然后答案就是max{i-last}，之前没有这个标记wa了一次。

代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;const int MAXN = 1000010;int num[MAXN];int q1[MAXN], q2[MAXN];int main(){ int n, m, k; while(~scanf("%d%d%d", &n, &m, &k)){ for(int i = 0; i < n; i++) scanf("%d", num + i); int h1 = 0, t1 = -1, h2 = 0, t2 = -1; int ans = 0; int last = -1; for(int i = 0; i < n; i++){ while(h1 <= t1 && num[q1[t1]] > num[i]) t1--; while(h2 <= t2 && num[q2[t2]] < num[i]) t2--; q1[++t1] = i; q2[++t2] = i; while(h1 <= t1 && h2 <= t2 && num[q2[h2]] - num[q1[h1]] > k){ if(q1[h1] < q2[h2]) last = q1[h1++]; else last = q2[h2++]; } if(h1 <= t1 && h2 <= t2 && num[q2[h2]] - num[q1[h1]] >= m){ ans = max(ans, i - last); } } printf("%d\n", ans); } return 0;}