介绍单调队列不错的博客:点击打开链接
还有这题输出用printf G++交会T,C++就能过.....而用cout两个都能过..好坑....
代码:
#include<cstdio>
using namespace std;
const int maxn = 1e6+5;
int n, k, ansMin[maxn], ansMax[maxn], queMax[maxn], queMin[maxn], val[maxn];
int main(void)
{
while(~scanf("%d%d", &n, &k))
{
int headMin = 0, tailMin = 0, headMax = 0, tailMax = 0;
for(int i = 0; i < n; i++)
{
while(headMin < tailMin && queMin[headMin] <= i-k) headMin++;
while(headMax < tailMax && queMax[headMax] <= i-k) headMax++;
scanf("%d", &val[i]);
while(headMin < tailMin && val[queMin[tailMin-1]] >= val[i]) tailMin--; tailMin++;
while(headMax < tailMax && val[queMax[tailMax-1]] <= val[i]) tailMax--; tailMax++;
queMin[tailMin-1] = queMax[tailMax-1] = i;
ansMin[i] = val[queMin[headMin]];
ansMax[i] = val[queMax[headMax]];
}
for(int i = k-1; i < n; i++) printf("%d%c", ansMin[i], i==n-1 ? '\n' : ' ');
for(int i = k-1; i < n; i++) printf("%d%c", ansMax[i], i==n-1 ? '\n' : ' ');
}
return 0;
}
Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 56208 | Accepted: 16159 | |
| Case Time Limit: 5000MS | ||
Description
An array of size
n ≤ 10
6 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.
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 |
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 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
本文介绍了一种使用单调队列求解滑动窗口中最小值和最大值的问题,并提供了一个具体的C++实现示例。该算法适用于数组长度不超过10^6的情况,能够高效地解决滑动窗口问题。

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