BZOJ 2096: [Poi2010]Pilots|单调队列

维护一个最大值的单调递减序列
维护一个最小值得单调递增序列

#include<set>#include<map>#include<ctime>#include<queue>#include<cmath>#include<cstdio>#include<vector>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#define T 3000006#define MX 1e9#define pa pair<int,int>using namespace std;priority_queue<pa,vector<pa>,greater<pa> >q;struct D{int x,pos;}mn[T],mx[T];int K,n,ans=0;int sc(){    int i=0,f=1; char c=getchar();    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9')i=i*10+c-'0',c=getchar();    return i*f;}int main(){    int ln=1,rn=0,lx=1,rx=0,L=0;    K=sc(),n=sc();    for(int i=1;i<=n;i++)    {        int x=sc();        while(rx>=lx&&mx[rx].x<=x)rx--;        while(rn>=ln&&mn[rn].x>=x)rn--;        mx[++rx]=mn[++rn]=(D){x,i};        while(mx[lx].x-mn[ln].x>K)        {            if(mx[lx].pos<mn[ln].pos)L=mx[lx].pos,lx++;            else L=mn[ln].pos,ln++;        }        ans=max(ans,i-L);    }    cout<<ans;    return 0;}