POJ2823——Sliding Window 单调队列入门
来源:互联网 发布:beta软件计划 编辑:程序博客网 时间:2024/06/04 19:00
Sliding Window
Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 40596 Accepted: 11992Case Time Limit: 5000MS
Description
An array of size n ≤ 106 is given to you. There is a sliding window of sizek which is moving from the very left of the array to the very right. You can only see thek numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.Window position Minimum value Maximum value [1 3 -1] -3 5 3 6 7 -13 1 [3 -1 -3] 5 3 6 7 -33 1 3 [-1 -3 5] 3 6 7 -35 1 3 -1 [-3 5 3] 6 7 -35 1 3 -1 -3 [5 3 6] 7 36 1 3 -1 -3 5 [3 6 7]37
The array is [1 3 -1 -3 5 3 6 7], and k is 3.
Your task is to determine the maximum and minimum values in the sliding window at each position.
Input
The input consists of two lines. The first line contains two integersn and k which are the lengths of the array and the sliding window. There aren integers in the second line.
Output
There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.
Sample Input
8 31 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 33 3 5 5 6 7
Source
以前做这题貌似用的RMQ,最近由于dp的需要,特意学了下单调队列
单调队列是一种支持队尾入队,两端可以出队的队,一般会维护一个索引值和value值
拿这题来说,我们以求最大为例,先把前k个入队,当然需要保证队列里是递减的,这样最大值才是队头,所以每次入队时,都要和队尾去比较,加入队尾太小,那么就要把队尾元素出队,直到可以把元素放进去
当然假如队头元素的索引值已经不在此时要求的窗口里,那么也要出队,每一个元素都在第i - k+1到第i个窗口,以此来判断队头是否要出队
由于这种单调性质,所以单调队列可以用来优化dp的转移
#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 1000010;int a[N], b[N];int arr[N];int x[N], y[N];int main(){int n, k;while(~scanf("%d%d", &n, &k)){for (int i = 1; i <= n; ++i){scanf("%d", &arr[i]);}if (k > n){k = n;}int front1, rear1, front2, rear2;int c1 = 0, c2 = 0;front1 = front2 = rear1 = rear2 = 0;for (int i = 1; i < k; ++i){while (front1 < rear1 && arr[a[rear1 - 1]] <= arr[i]){rear1--;}a[rear1++] = i;while (front2 < rear2 && arr[b[rear2 - 1]] >= arr[i]){rear2--;}b[rear2++] = i;}for (int i = k; i <= n; ++i){while (front1 < rear1 && arr[a[rear1 - 1]] <= arr[i]){rear1--;}a[rear1++] = i;if (a[front1] < i - k + 1){front1++;}x[++c1] = arr[a[front1]];while (front2 < rear2 && arr[b[rear2 - 1]] >= arr[i]){rear2--;}b[rear2++] = i;if (b[front2] < i - k + 1){front2++;}y[++c2] = arr[b[front2]];}printf("%d", y[1]);for (int i = 2; i <= c2; ++i){printf(" %d", y[i]);}printf("\n");printf("%d", x[1]);for (int i = 2; i <= c1; ++i){printf(" %d", x[i]);}printf("\n");}return 0;}
0 0
- POJ2823——Sliding Window 单调队列入门
- POJ2823.Sliding Window——单调队列
- ACM-单调队列之Sliding Window——poj2823
- Sliding Window poj2823--单调队列
- POJ2823 Sliding Window单调队列
- POJ2823 Sliding Window(单调队列)
- POJ2823 Sliding Window(单调队列)
- 单调队列 poj2823 Sliding Window
- POJ2823 Sliding Window(单调队列)
- 【POJ2823】Sliding Window-单调队列
- poj2823 Sliding Window 单调队列
- poj2823 Sliding Window 单调队列
- [POJ2823]Sliding Window(单调队列)
- [POJ2823]Sliding Window(单调队列)
- POJ2823 Sliding Window【单调队列】
- 单调队列--poj2823 Sliding Window
- POJ2823 Sliding Window,手工实现单调队列
- POJ2823:Sliding Window(单调队列||线段树)
- Spring Mvc 的核心组件
- Myeclipse30天试用期过了之后如何破解
- 安卓第四课思考:加上时间限制
- scandir 函数
- 5 Open Source Business Intelligence Tools[翻译]
- POJ2823——Sliding Window 单调队列入门
- vim多窗口编辑
- UVA10183 - How Many Fibs?(java大数+二分)
- poj 1006 Biorhythms
- C#MD5加密16进制写法
- 如何更改MFC程序图标
- 几种平衡树的总结
- brk(), sbrk() 用法详解 (转)
- wp开发 progress bar的制作