单调队列入门

Posted 2020-09-11 R o b i n

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

 

 1 //O(n)
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 const int N = 1000 + 5;
 5 int a[N];
 6 
 7 struct MonotonicQueue {
 8     int minv[N], maxv[N], q[N], id[N], n, k;
 9 
10     void init(int n, int k) {
11         this->n = n; this->k = k;
12     }
13 
14     void getmax(int a[]) {
15         int head = 0, tail = 0;
16         for (int i = 0; i < k-1; ++i) {
17             while (head < tail && q[tail-1] <= a[i]) --tail;
18             id[tail] = i;
19             q[tail++] = a[i];
20         }
21         for (int i = k-1; i < n; ++i) {
22             while (head < tail && q[tail-1] <= a[i]) --tail;
23             id[tail] = i;
24             q[tail++] = a[i];
25             while (id[head] < i-k+1) ++head;
26             for (int j = head; j < tail; ++j) cout << q[j] << " ";
27             cout << endl;
28             maxv[i-k+1] = q[head];
29         }
30     }
31 
32     void getmin(int a[]) {
33         int head = 0, tail = 0;
34         for (int i = 0; i < k-1; ++i) {
35             while (head < tail && q[tail-1] >= a[i]) --tail;
36             id[tail] = i;
37             q[tail++] = a[i];
38         }
39         for (int i = k-1; i < n; ++i) {
40             while (head < tail && q[tail-1] >= a[i]) --tail;
41             id[tail] = i;
42             q[tail++] = a[i];
43             while (id[head] < i-k+1) ++head;
44             minv[i-k+1] = q[head];
45         }
46     }
47 
48 }mq;
49 
50 int main() {
51     int n, k;
52     while (~scanf("%d%d", &n, &k)) {
53         mq.init(n, k);
54         for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
55         mq.getmax(a); mq.getmin(a);
56         for (int i = 0; i < n-k+1; ++i) 
57             printf("%d %d\\n", mq.minv[i], mq.maxv[i]);
58     }
59 
60     return 0;
61 }