单调队列入门

给你一个长度为N的数组，一个长为K的滑动的窗体从最左移至最右端，你只能见到窗口的K个数，每次窗体向右移动一位，如下表：



你的任务是找出窗口在各位置时的max value,min value.

单调队列的入门题，单调队列我的理解有对时间和大小都单调，只会在队列两端进行操作（队尾前移看大小，队头后移看时间），不会有元素插入到之前的两元素之间，因此每次都能保证队头元素是符合要求的

//O(n)
#include<bits/stdc++.h>
using namespace std;
const int N = 1000 + 5;
int a[N];

struct MonotonicQueue {
    int minv[N], maxv[N], q[N], id[N], n, k;

    void init(int n, int k) {
        this->n = n; this->k = k;
    }

    void getmax(int a[]) {
        int head = 0, tail = 0;
        for (int i = 0; i < k-1; ++i) {
            while (head < tail && q[tail-1] <= a[i]) --tail;
            id[tail] = i;
            q[tail++] = a[i];
        }
        for (int i = k-1; i < n; ++i) {
            while (head < tail && q[tail-1] <= a[i]) --tail;
            id[tail] = i;
            q[tail++] = a[i];
            while (id[head] < i-k+1) ++head;
            for (int j = head; j < tail; ++j) cout << q[j] << " ";
            cout << endl;
            maxv[i-k+1] = q[head];
        }
    }

    void getmin(int a[]) {
        int head = 0, tail = 0;
        for (int i = 0; i < k-1; ++i) {
            while (head < tail && q[tail-1] >= a[i]) --tail;
            id[tail] = i;
            q[tail++] = a[i];
        }
        for (int i = k-1; i < n; ++i) {
            while (head < tail && q[tail-1] >= a[i]) --tail;
            id[tail] = i;
            q[tail++] = a[i];
            while (id[head] < i-k+1) ++head;
            minv[i-k+1] = q[head];
        }
    }

}mq;

int main() {
    int n, k;
    while (~scanf("%d%d", &n, &k)) {
        mq.init(n, k);
        for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
        mq.getmax(a); mq.getmin(a);
        for (int i = 0; i < n-k+1; ++i) 
            printf("%d %d
", mq.minv[i], mq.maxv[i]);
    }

    return 0;
}