P3512 [POI2010]PIL-Pilots

P3512 [POI2010]PIL-Pilots
我一开始打的O（n^2）（最坏情况）的算法.枚举区间长度。60分

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstring>
#define len 3000010
#define inf 2147483647
#define For(i,a,b) for(register int i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()
//by war
//2017.10.9
using namespace std;
int n,c;
int a[len];
int qz[len];
int qj[len];
int lz,rz,lj,rj;
void in(int &x)
{
    char c=g();x=0;
    while(c<'0'||c>'9')c=g();
    while(c<='9'&&c>='0')x=x*10+c-'0',c=g();
}

void o(int x)
{
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main()
{
     in(c),in(n);
     For(i,1,n)
     in(a[i]);
     for(register int k=n;k>=1;k--)
     {
         lz=1,rz=0,lj=1,rj=0;
         For(i,1,n)
         {
             
          while(lj<=rj&&a[qj[rj]]<=a[i])rj--;
         qj[++rj]=i;
         while(qj[rj]-qj[lj]>=k)lj++;
         
         while(lz<=rz&&a[qz[rz]]>=a[i])rz--;
         qz[++rz]=i;
         while(qz[rz]-qz[lz]>=k)lz++;
         
         if(i>=k)
         {
             if(a[qj[lj]]-a[qz[lz]]<=c)
             {
              o(k);
             exit(0);    
            }
         }    
        }
     }
     return 0;
}

正解不用枚举区间长度，动态维护+更新答案,O(n).

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<cstring>
#define len 3000010
#define inf 2147483647
#define For(i,a,b) for(register int i=a;i<=b;i++)
#define p(a) putchar(a)
#define g() getchar()
//by war
//2017.10.9
using namespace std;
int n,c;
int a[len];
int qz[len];
int qj[len];
int lz,rz,lj,rj;
int ans,last;
void in(int &x)
{
    char c=g();x=0;
    while(c<'0'||c>'9')c=g();
    while(c<='9'&&c>='0')x=x*10+c-'0',c=g();
}

void o(int x)
{
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main()
{
     in(c),in(n);
     For(i,1,n)
     in(a[i]);
         lz=1,rz=0,lj=1,rj=0;
         For(i,1,n)
         {
             
          while(lj<=rj&&a[qj[rj]]<=a[i])rj--;
          while(lz<=rz&&a[qz[rz]]>=a[i])rz--;
         qj[++rj]=i;qz[++rz]=i;
         while(a[qj[lj]]-a[qz[lz]]>c)
         if(qj[lj]<qz[lz])
         last=qj[lj],lj++;
         else
         last=qz[lz],lz++;
         ans=max(ans,i-last);
        }
        o(ans);
     return 0;
}