Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 55520 | Accepted: 15971 | |
| Case 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.
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
题目解析：
题目大意是给你一个长度为n的序列，长度为k的窗口，窗口不断滑动，求每次移动时窗口内的最大值和最小值。
据说是单调队列裸题。我用G++语言交就会TLE，用C++交就A掉了…
我们开两个双端队列来分别维护一个单调递减的队列和一个单调递增的队列。以维护递增队列为例，因为他前面出现的值小，所以肯定不会是最优解，就可以把它pop掉。这是正确的。每次读入时我们把它push进去，push的时候要不断维护更新这个队列，同时我们要判断一下队列有没有超出长度限制。这样的队头就是这个窗口中最大或最小的值了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int numm[2333333];
struct qer
{
int val, num;
};
deque <qer> q1;
deque <qer> q2;
void push1(int x, int y)
{
while(!q1.empty() && q1.back().val <= x)
q1.pop_back();
qer a;
a.val = x;
a.num = y;
q1.push_back(a);
}
void push2(int x, int y)
{
while(!q2.empty() && q2.back().val >= x)
q2.pop_back();
qer a;
a.val = x;
a.num = y;
q2.push_back(a);
}
int main()
{
int n, k;
scanf("%d%d", &n, &k);
for(int i = 1;i <= n;i ++)
{
scanf("%d", &numm[i]);
push2(numm[i], i);
while(i-k >= q2.front().num)
q2.pop_front();
if(i >= k)
printf("%d ", q2.front().val);
}
puts("");
for(int i = 1;i <= n;i ++)
{
push1(numm[i], i);
while(i-k >= q1.front().num)
q1.pop_front();
if(i >= k)
printf("%d ", q1.front().val);
}
return 0;
}