【BZOJ】【P2096】【Poi2010】【Pilots】【题解】【二分+单调队列】

来源：互联网发布：淘宝淘口令在哪看编辑：程序博客网时间：2024/06/05 03:55

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=2096

答案满足二分性质……二分+判定

nlogn好像很慢……

Code:

#include<bits/stdc++.h>using namespace std;const int maxn=3e6+5;int k,n;int a[maxn];deque<int>Min,Max;int main(){scanf("%d%d",&k,&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);int l=1,r=n+1;while(l<r){int mid=(l+r)>>1,ok=0;Min.clear();Max.clear();for(int i=1;i<=mid;i++){while(Max.size()&&a[Max.back()]<=a[i])Max.pop_back();while(Min.size()&&a[Min.back()]>=a[i])Min.pop_back();Max.push_back(i);Min.push_back(i);}if(a[Max.front()]-a[Min.front()]<=k){l=mid+1;ok=1;continue;}for(int i=1;i+mid<=n;i++){while(Max.size()&&Max.front()<=i)Max.pop_front();while(Min.size()&&Min.front()<=i)Min.pop_front();while(Max.size()&&a[Max.back()]<=a[i+mid])Max.pop_back();while(Min.size()&&a[Min.back()]>=a[i+mid])Min.pop_back();Max.push_back(i+mid);Min.push_back(i+mid);if(a[Max.front()]-a[Min.front()]<=k){l=mid+1;ok=1;break;}}if(!ok)r=mid;}cout<<l-1<<endl;return 0;}

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