HDU3530（单调队列）

Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6679 Accepted Submission(s): 2243

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

题意：给你N个数，让你找出一个最长的连续子序列，这个子序列的最大值和最小值的差值介于m,k之间。

解题思路：用两个单调队列，一个维护最大值，一个维护最小值，当最大值与最小值的差值大于k时，考虑把最大值变小，或者把最小值变大，具体看最大值与最小值的位置关系。、

#include<bits/stdc++.h>using namespace std;const int maxn = 1e5 + 10;int qmax[maxn];//维护最大值的队列int qmin[maxn];//维护最小值的队列int lmax, rmax;int lmin, rmin;//l 表示队尾，r表示队首int n, m, k;int a[maxn];int ans;void init(){    lmax = lmin = 0;    rmin = rmax = 0;    ans = 0;}int main(){    while(~scanf("%d%d%d", &n, &m, &k))    {        init();        for(int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);        }        int l1 = 0;        int l2 = 0;        bool flag = false;        for(int i = 1; i <= n; i++)        {            while(rmax > lmax && a[qmax[rmax - 1]] <= a[i]) rmax--;            qmax[rmax++] = i;            while(rmin > lmin && a[qmin[rmin - 1]] >= a[i]) rmin--;            qmin[rmin++] = i;            while(a[qmax[lmax]] - a[qmin[lmin]] > k)            {                if(qmax[lmax] > qmin[lmin]) l2 = qmin[lmin++];                else l1 = qmax[lmax++];            }            if(a[qmax[lmax]] - a[qmin[lmin]] >= m && a[qmax[lmax]] - a[qmin[lmin]] <= k)            {                ans = max(ans,i - max(l1,l2));                flag = true;            }        }        if(flag)        printf("%d\n", ans);        else printf("0\n");    }    return 0;}