#POJ2823#Sliding Window (单调队列基础)
来源:互联网 发布:centos配置内网ip 编辑:程序博客网 时间:2024/06/11 01:00
Sliding Window
Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 60091 Accepted: 17230Case Time Limit: 5000MS
Description
An array of size n ≤ 106 is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k 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 integers n and k which are the lengths of the array and the sliding window. There are n 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
单调队列基础题
#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;const int Max = 1000005;int N, K, fro, back;int A[Max];struct queue{int v, num;queue(){}queue(int a, int b){v = a, num = b;}}Q[Max];void getint(int&num){ char c;int flag=1;num=0; while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1; while(c>='0'&&c<='9'){num=num*10+c-48;c=getchar();} num*=flag;}int main(){getint(N), getint(K);bool flg = 0;fro = 1, back = 0;for(int i = 1; i <= N; ++ i){getint(A[i]);while(i - Q[fro].num + 1 > K && fro <= back)++ fro;while(fro <= back && A[i] <= Q[back].v)-- back;Q[++ back] = queue(A[i], i);if(i >= K){if(flg)putchar(32);printf("%d", Q[fro].v);flg = 1;}}putchar(10);fro = 1, back = 0;flg = 0;for(int i = 1; i <= N; ++ i){while(i - Q[fro].num + 1 > K && fro <= back)++ fro;while(fro <= back && A[i] >= Q[back].v)--back;Q[++back] = queue(A[i], i);if(i >= K){if(flg)putchar(32);printf("%d", Q[fro].v);flg = 1;}}putchar(10);return 0;}
阅读全文
0 0
- #POJ2823#Sliding Window (单调队列基础)
- 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)
- poj2823--Sliding Window--线段树||单调队列
- POJ2823.Sliding Window——单调队列
- ORA-01704: 文字字符串过长成功解决
- 数据库面试知识
- TweenJS 一个简单但强大的渐变界面
- setjmp.h学习笔记
- Python程序 Catmouseme
- #POJ2823#Sliding Window (单调队列基础)
- JavaScript原型的缺点及改进
- java关键字
- 搭建线程池需要考虑的因素
- 关于synchronized 特别需要强调的一个点!!
- 一共有20级楼梯,每次可以上1级或2级,登上第20级一共有多少种上法
- 基于STM32wifi小车制作(二)-电源设计
- jsp forward指令
- Radar Installation(POJ-1328)