這個裸題，滑動窗口求最大最小值，單調隊列來兩邊，一次單調遞增q[s]就是最小值，一次單調遞減q[s]就是最大值

cin會超時，解除同步也沒用。。。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include
 
<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-9;
const int N=1000000+10,maxn=500+100,inf=0x3f3f3f
 
;

int a[N],q[N];
int minn[N],maxx[N];
int main()
{
 /*   ios::sync_with_stdio(false);
    cin.tie(0);*/
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        int s=0,t=0;
        for(int i=0;i<n;i++)
        {
            while(s<t&&a[i]>a[q[t-1
 
]])t--;
            q[t++]=i;
            if(s<t&&q[t-1]-q[s]>=k)s++;
        /*    for(int j=s;j<t;j++)cout<<a[q[j]]<<" ";
            cout<<endl;*/
            minn[i]=a[q[s]];
        }
        s=0,t=0;
        for(int i=0;i<n;i++)
        {
            while(s<t&&a[i]<a[q[t-1]])t--;
            q[t++]=i;
            if(s<t&&q[t-1]-q[s]>=k)s++;
          /*  for(int j=s;j<t;j++)cout<<a[q[j]]<<" ";
            cout<<endl;*/
            maxx[i]=a[q[s]];
        }
        for(int i=k-1;i<n;i++)
            printf("%d%c",maxx[i],i==n-1?‘\n‘:‘ ‘);
        for(int i=k-1;i<n;i++)
            printf("%d%c",minn[i],i==n-1?‘\n‘:‘ ‘);
    }
    return 0;
}

poj2823單調隊列