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Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 40565 | Accepted: 11982 | |
| 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 knumbers 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
用一个长度为k的窗在整数数列上移动，求窗里面所包含的数的最小值最大值。
单调队列定义：
a)从队头到队尾，元素在我们所关注的指标下是递减的（严格递减，而不是非递增），比如查询如果每次问的是窗口内的最小值，那么队列中元素从左至右就应该递增，如果每次问的是窗口内的最大值，则应该递减，依此类推。这是为了保证每次查询只需要取队头元素。
b)从队头到队尾，元素对应的时刻（此题中是该元素在数列a中的下标）是递增的，但不要求连续，这是为了保证最左面的元素总是最先过期，且每当有新元素来临的时候一定是插入队尾。
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int que[1111111],num[1111111],maxn[1111111],minn[1111111],index[1111111];
int n,k;
void getmin()
{
int fr=1,ed=0,i;
for(i=0;i<k-1;i++)
{
while(fr<=ed&&que[ed]>=num[i])
ed--;
que[++ed]=num[i];
index[ed]=i;
}
for(;i<n;i++)
{
while(fr<=ed&&que[ed]>=num[i])
ed--;
que[++ed]=num[i];
index[ed]=i;
while(index[fr]<i-k+1)
{
fr++;
}
minn[i-k+1]=que[fr];
}
}
void getmax()
{
int fr=1,ed=0,i;
for(i=0;i<k-1;i++)
{
while(fr<=ed&&que[ed]<=num[i])
ed--;
que[++ed]=num[i];
index[ed]=i;
}
for(;i<n;i++)
{
while(fr<=ed&&que[ed]<=num[i])
ed--;
que[++ed]=num[i];
index[ed]=i;
while(index[fr]<i-k+1)
{
fr++;
}
maxn[i-k+1]=que[fr];
}
}
int main()
{
//freopen("acm.txt","r",stdin);
scanf("%d %d",&n,&k);
for(int i=0;i<n;i++)
{
scanf("%d",&num[i]);
}
getmin();
getmax();
for(int i=0;i<n-k+1;i++)
{
if(i<n-k)
printf("%d ",minn[i]);
else
printf("%d\n",minn[i]);
}
for(int i=0;i<n-k+1;i++)
{
if(i<n-k)
printf("%d ",maxn[i]);
else
printf("%d\n",maxn[i]);
}
return 0;
}