单调队列1003 HDU 3530 Subsequence

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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.

题意:
给一个长度为n的序列,从中找到一个最长的连续的子序列满足
最大的元素和最小的元素的差>=m&&<=k
思路:
维护两个单调队列,维护的时候是维护最大值减去最小值<=ki就行了

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>#include<queue>#include<stack>#include<string>#include<vector>#include<map>#include<set>using namespace std;#define lowbit(x) (x&(-x))typedef long long LL;const int maxn = 100005;const int inf=(1<<28)-1;deque<pair<int,int> >QueMax,QueMin;int main(){    int n,k,m;    while(~scanf("%d%d%d",&n,&m,&k))    {        int ans=0;        QueMax.clear();        QueMin.clear();        int MinPos=-1;        for(int i=0;i<n;++i)        {            int x;            scanf("%d",&x);            while(!QueMax.empty()&&QueMax.back().first<x) QueMax.pop_back();            while(!QueMin.empty()&&QueMin.back().first>x) QueMin.pop_back();            QueMax.push_back(make_pair(x,i));            QueMin.push_back(make_pair(x,i));            while(!QueMax.empty()&&!QueMin.empty()&&QueMax.front().first-QueMin.front().first>k)            {                if(QueMax.front().second>QueMin.front().second)                {                    MinPos=QueMin.front().second;                    QueMin.pop_front();                }                else                {                    MinPos=QueMax.front().second;                    QueMax.pop_front();                }            }            if(QueMax.front().first-QueMin.front().first>=m)            ans=max(ans,i-MinPos);        }        printf("%d\n",ans);    }    return 0;}

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