POJ

分别维护一个单调递减的队列和单调递增队列就可以了，但是这题莫名的坑，要用C++提交卡编译器的时间才能AC，真的是日了狗了，以后就用C++提交好了，不用别的了。

#include <iostream>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <sstream>
#include <time.h>
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define lson l,m,rt*2
#define rson m+1,r,rt*2+1
#define mt(A,B) memset(A,B,sizeof(A))
using namespace std;
typedef long long LL;
//const double PI = acos(-1);
const int N=1e6+10;
const LL mod=1e9+7;
const int inf = 0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3fLL;
int a[N],MI[N],MA[N];
struct node
{
    int x,y;
}Q[N];//第一位是下标，第二位是数据
int main()
{
#ifdef Local
    freopen("data.txt","r",stdin);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    int n,k,w=0;
    register int head=0,tail=0;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)scanf("%d",&a[i]);
    for(int i=0;i<n;i++)
    {
        while(tail-head>0&&Q[head+1].x<i-k+1)head++;
        while(tail>head&&Q[tail].y>=a[i])tail--;
        tail++;
        Q[tail].y=a[i];
        Q[tail].x=i;
        if(i>=k-1)MI[w++]=Q[head+1].y;
    }
    head=0;tail=0;w=0;
    for(int i=0;i<n;i++)
    {
        while(tail-head>0&&Q[head+1].x<i-k+1)head++;
        while(tail>head&&Q[tail].y<=a[i])tail--;
        tail++;
        Q[tail].y=a[i];
        Q[tail].x=i;
        if(i>=k-1)MA[w++]=Q[head+1].y;
    }
    for(int i=0;i<w;i++)
    {
        if(i==w-1)printf("%d
",MI[i]);
        else printf("%d ",MI[i]);
    }
    for(int i=0;i<w;i++)
    {
        if(i==w-1)printf("%d
",MA[i]);
        else printf("%d ",MA[i]);
    }
#ifdef Local
    cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
#endif
}