POJ 题目2823 Sliding Window(单调队列求定长区间最大值)
2015-09-25 18:57
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Sliding Window
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.
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
Sample Output
Source
POJ Monthly--2006.04.28, Ikki
来了机房就开始睡。。睡醒来一发,
开始做过这题,用rmq做的,又用单调队列试了一发,时间比rmq省一些,内存省了不少
rmq做法:http://blog.csdn.net/yu_ch_sh/article/details/47083283
ac代码
Time Limit: 12000MS | Memory Limit: 65536K | |
Total Submissions: 48701 | Accepted: 14078 | |
Case Time Limit: 5000MS |
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 | -1 | 3 |
1 [3 -1 -3] 5 3 6 7 | -3 | 3 |
1 3 [-1 -3 5] 3 6 7 | -3 | 5 |
1 3 -1 [-3 5 3] 6 7 | -3 | 5 |
1 3 -1 -3 [5 3 6] 7 | 3 | 6 |
1 3 -1 -3 5 [3 6 7] | 3 | 7 |
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 3 1 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 3 3 3 5 5 6 7
Source
POJ Monthly--2006.04.28, Ikki
来了机房就开始睡。。睡醒来一发,
开始做过这题,用rmq做的,又用单调队列试了一发,时间比rmq省一些,内存省了不少
rmq做法:http://blog.csdn.net/yu_ch_sh/article/details/47083283
ac代码
Problem: 2823 User: kxh1995 Memory: 4048K Time: 5282MS Language: C++ Result: Accepted
#include<iostream> #include<string.h> #include<stdio.h> #include<algorithm> using namespace std; #define N 1000010 int a ,q ; int main() { int n,k; while(scanf("%d%d",&n,&k)!=EOF) { int i,j; for(i=1;i<=n;i++) { scanf("%d",&a[i]); } int front,rear; front=0; rear=-1; for(i=1;i<=k;i++) { for(;front<=rear&&a[i]<=a[q[rear]];rear--); q[++rear]=i; } printf("%d",a[q[front]]); for(j=1;i<=n;i++,j++) { if(front<=rear&&q[front]<=j) front++; for(;front<=rear&&a[i]<=a[q[rear]];rear--); q[++rear]=i; printf(" %d",a[q[front]]); } printf("\n"); front=0; rear=-1; for(i=1;i<=k;i++) { for(;front<=rear&&a[i]>=a[q[rear]];rear--); q[++rear]=i; } printf("%d",a[q[front]]); for(j=1;i<=n;i++,j++) { if(front<=rear&&q[front]<=j) front++; for(;front<=rear&&a[i]>=a[q[rear]];rear--); q[++rear]=i; printf(" %d",a[q[front]]); } printf("\n"); } }
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