HDU 3530 Subsequence（单调队列）

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7230    Accepted Submission(s): 2448

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
  

 

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU 

 

题意：

找出最大的区间长度，这个区间里的最大值和最小值的差值不能小于m，不能大于k。

POINT：

维护两个单调队列。当差值大于k时，从最大值和最小值中删去靠前的那一个。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
using namespace std;
const int N = 100100+4;
int a[N];
deque<int> q1;
int ans;
deque<int> q2;
int main()
{
    int n,m,k;
    while(~scanf("%d %d %d",&n,&m,&k))
    {
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        int last=0;//注意初值。
        while(!q1.empty()) q1.pop_front();
        while(!q2.empty()) q2.pop_front();
        for(int i=1;i<=n;i++)
        {
            while(!q1.empty()&&a[i]>=a[q1.back()]) q1.pop_back();
            while(!q2.empty()&&a[i]<=a[q2.back()]) q2.pop_back();
            q1.push_back(i);
            q2.push_back(i);
            while(!q1.empty()&&!q2.empty()&&a[q1.front()]-a[q2.front()]>k)
            {
                if(q1.front()>q2.front())
                {
                    last=q2.front();
                    q2.pop_front();
                }
                else
                {
                    last=q1.front();
                    q1.pop_front();
                }
            }
            if(!q1.empty()&&!q2.empty()&&a[q1.front()]-a[q2.front()]>=m)
            {
                ans=max(ans,i-last);
            }
        }
        printf("%d\n",ans);
    }
}