模板：单调队列

首先给出自己封装好的模板吧。

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template<typename T, unsigned int N, bool (*func)(T, T)>
class MQueue {
private :
    pair<T , int >q[N];
    int hd, tl, len;
public :
    inline void clear(int len) { this->len = len; hd = 0, tl = -1; }
    inline int size() { return tl - hd + 1; }
    inline bool empty() { return hd > tl; }
    inline void push(pair<T, int> ele) {
        while(!empty() && (*func)(ele.first, q[tl].first)) tl--;
        q[++tl] = ele;
        while(!empty() && q[hd].second <= ele.second - len) hd++;
    }
    inline T front() { return q[hd].first; }
};

代码：模板试用

这题真的没有难度。

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// #pragma GCC optimize(2)
// #pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef __int128 int128;

namespace FastIO 
{
    // char buf[1 << 20], *p1, *p2;
    // #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? EOF : *p1++)
    template<typename T> inline T read(T& x) {
        x = 0;
        int f = 1;
        char ch;
        while (!isdigit(ch = getchar())) if (ch == '-') f = -1;
        while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
        x *= f;
        return x;
    }
    template<typename T, typename... Args> inline void read(T& x, Args &...x_) {
        read(x);
        read(x_...);
        return;
    }
    inline ll read() {
        ll x;
        read(x);
        return x;
    }
};
using namespace FastIO;

template<typename T, unsigned int N, bool (*func)(T, T)>
class MQueue {
private :
    pair<T , int >q[N];
    int hd, tl, len;
public :
    inline void clear(int len) { this->len = len; hd = 0, tl = -1; }
    inline int size() { return tl - hd + 1; }
    inline bool empty() { return hd > tl; }
    inline void push(pair<T, int> ele) {
        while(!empty() && (*func)(ele.first, q[tl].first)) tl--;
        q[++tl] = ele;
        while(!empty() && q[hd].second <= ele.second - len) hd++;
    }
    inline T front() { return q[hd].first; }
};

bool cmp1(int a, int b) { return a < b; }
bool cmp2(int a, int b) { return a > b; }

MQueue<int, 1000010, cmp1>q1;
MQueue<int, 1000010, cmp2>q2;

int n, m;
int a[1000010];

inline void Input() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
}

inline void Work() {
    q1.clear(m), q2.clear(m);
    for(int i = 1; i <= n; i++) {
        q1.push(make_pair(a[i], i));
        if(i >= m) printf("%d ", q1.front());
    }
    printf("\n");
    for(int i = 1; i <= n; i++) {
        q2.push(make_pair(a[i], i));
        if(i >= m) printf("%d ", q2.front());
    }
}

int main() {
    int T = 1;
    while(T--) {
        Input();
        Work();
    }
    return 0;
}