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来源：互联网发布：中国骨科手术量数据编辑：程序博客网时间：2024/06/17 12:11

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：

这里写图片描述

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 }复制代码

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