poj2823 Sliding Window

最新推荐文章于 2021-05-22 16:00:30 发布

elijahqi

最新推荐文章于 2021-05-22 16:00:30 发布

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分类专栏：单调队列

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本文介绍了一种解决滑动窗口中寻找最大值与最小值问题的高效算法。通过使用单调队列，该算法能够在O(n)的时间复杂度内完成计算。文章详细解释了如何维护递增和递减序列来更新窗口内的极值，并提供了完整的C++实现代码。

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http://www.elijahqi.win/archives/983
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 -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
单调队列的题咳咳告诉自己不要浮躁，不要急于追求结果

这题注意下poj上不知道为什么c++编译才行，否则会tle

维护最大值&最小值，我们设立两个单调队列

并且每个队列都是两个指针

先说最小值：我们维护递增的序列

当左指针所指位置存的值不在我们的框内，我们选择右移左指针，当我队列里的值大于我新入队的值，我把右指针回移，相当于弹出

再说最大值：我们维护递减的序列

当做指针所指位置存的值不在我们的框内，我们选择右移左指针，这次，当队列中的值小于我新入队的值的话，将右指针回移，弹出我们的不满足条件的数据

#include<cstdio>
#define N 1100000
inline int read(){
    int x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int up[N],down[N],ans1[N],ans2[N],a[N],n,k,r1,r2; 
int main(){
    freopen("poj2823.in","r",stdin);
    while (~scanf("%d%d",&n,&k)){
        for (int i=1;i<=n;++i) a[i]=read();
        int l1=1,l2=1;r1=r2=0;
        for (int i=1;i<=n;++i){
            while (l1<=r1&&down[l1]<=i-k) ++l1;
            while (l2<=r2&&up[l2]<=i-k) ++l2;
            while (l1<=r1&&a[i]<a[down[r1]]) r1--;down[++r1]=i;
            while (l2<=r2&&a[i]>a[up[r2]]) r2--;up[++r2]=i;
            ans1[i]=a[down[l1]];ans2[i]=a[up[l2]];
        }
        for (int i=k;i<=n;++i) printf("%d ",ans1[i]);printf("\n");
        for (int i=k;i<=n;++i) printf("%d ",ans2[i]);
    }
    return 0;
}