hdu3530 单调队列

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7255    Accepted Submission(s): 2458

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 01 1 1 1 15 0 31 2 3 4 5

 

Sample Output

54
题意：
给你一串数字组成的数组，求最长的子串满足，字串的最大值减去子串的最小值大于等于m，小于等于k
单调队列的思想,i表示满足条件的子串的终点，一个单调递增的队列用来保存比i大的数，一个单调递减队列用来保存比i小的数字，对于每个i相当于枚举起点，找到合适的
实际上算法很巧秒的使每个位置只枚举一次
具体看程序，然后自己多模拟理解理解，就会豁然开朗
#include<cstdio>#include<string>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<map>#include<set>#include<stack>#define ll long long#define read(a) scanf("%d",&a);using namespace std;const int maxn=1e5+10;int a[maxn],qmax[maxn],qmin[maxn];int main(){freopen("test.txt","r",stdin);int n,m,k;while(~scanf("%d %d %d",&n,&m,&k)){int head1,tail1,head2,tail2;int last=0;int ans=0;head1=tail1=head2=tail2=0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);while(head1<tail1&&a[qmax[tail1-1]]<a[i])tail1--;while(head2<tail2&&a[qmin[tail2-1]]>a[i])tail2--;qmax[tail1++]=i;qmin[tail2++]=i;while(a[qmax[head1]]-a[qmin[head2]]>k){last=(qmax[head1]>qmin[head2]?qmin[head2++]:qmax[head1++]);}if(a[qmax[head1]]-a[qmin[head2]]>=m){ans=max(ans,i-last);}}printf("%d\n",ans);}return 0;}