洛谷 2032.扫描

思路：单调队列

就是一个板子题，直接上代码：

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<cmath> 
#include<vector>
#include<algorithm>
#include<stack>
#include<queue>
#include<deque>
#include <iomanip>
#include<sstream>
#include<numeric>
#include<map>
#include<limits.h>
#include<unordered_map>
#include<set>
#define int long long
#define MAX 2000200
#define inf 0x3f3f3f3f
#define _for(i,a,b) for(int i=a;i<(b);i++)
#define ALL(x) x.begin(),x.end()
using namespace std;
typedef pair<int, int> PII;
int n, m, k;
int counts=inf;
int dx[] = { 0,1,0,-1};
int dy[] = { 1,0,-1,0 };
int q[MAX];
int arr[MAX];
int b[MAX];
void get_max(int a[], int b[], int n, int qujian) {
    int front = 0;
    int rear = -1;
    for (int i = 0; i < n; i++) {
        if (front <= rear && q[front] + qujian <= i)front++;
        while (front <= rear && a[q[rear]] <= a[i])rear--;
        q[++rear] = i;
        if (i >= qujian - 1)
            b[i] = a[q[front]];
    }
}
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }
    get_max(arr, b, n, k);
    _for(i, k - 1, n)
        cout << b[i] << endl;
    return 0;
}