Presented written out of the tree line(Long live the tree line)
Build tree twice, once deposit a minimum, a maximum deposit
The last point limit card time 980ms
#include<bits/stdc++.h>
using namespace std;
const int N=1000000;
inline int Read()//快读
{
int x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
return x*f;
}
struct Tree
{
int l,r,dat;
Tree()
{
l=r=0;
dat=0;
}
}a[(N<<2)+10];
int num[N+10];
void Buildmin(int l,int r,int s)//建立存储最小值的线段树
{
a[s].l=l;
a[s].r=r;
if(l==r)
{
a[s].dat=num[l];
return;
}
int mid=(l+r)>>1;
Buildmin(l,mid,s<<1);
Buildmin(mid+1,r,s<<1|1);
a[s].dat=min(a[s<<1].dat,a[s<<1|1].dat);
}
void Buildmax(int l,int r,int s)//建立存储最大值的线段树
{
a[s].l=l;
a[s].r=r;
if(l==r)
{
a[s].dat=num[l];
return;
}
int mid=(l+r)>>1;
Buildmax(l,mid,s<<1);
Buildmax(mid+1,r,s<<1|1);
a[s].dat=max(a[s<<1].dat,a[s<<1|1].dat);
}
int querymax(int l,int r,int s)//查询区间[l,r]最大值
{
int ans=INT_MIN;
if(a[s].l>=l&&a[s].r<=r) return a[s].dat;
if(a[s<<1].r>=l) ans=max(ans,querymax(l,r,s<<1));
if(a[s<<1|1].l<=r) ans=max(ans,querymax(l,r,s<<1|1));
return ans;
}
int querymin(int l,int r,int s)//查询区间[l,r]最小值
{
int ans=INT_MAX;
if(a[s].l>=l&&a[s].r<=r) return a[s].dat;
if(a[s<<1].r>=l) ans=min(ans,querymin(l,r,s<<1));
if(a[s<<1|1].l<=r) ans=min(ans,querymin(l,r,s<<1|1));
return ans;
}
int main()
{
int n=Read(),k=Read();
int from=1,to=1+k-1;//from:窗口的左端点 to:窗口的右端点
for(int i=1;i<=n;i++) num[i]=Read();
Buildmin(1,n,1);//第一次建树，存最小值
while(to<=n)
{
printf("%d ",querymin(from,to,1));
from++;
to++;
}
puts("");
from=1,to=1+k-1;
Buildmax(1,n,1);//第二次建树，存最大值
while(to<=n)
{
printf("%d ",querymax(from,to,1));
from++;
to++;
}
return 0;
}