[ACM] poj 2823 Sliding Window （单调队列）

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 36315		Accepted: 10756
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.

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

 

 

解题思路：

 

[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

 

单调队列里存得是下标。这里为了好理解些，假设队列里存得是元素。

要求最大值，则维护一个单调递减的队列。队首即为答案。

1 3 -1 -3 5 3 6 7    K=3

准备插入1   队列为空， 插入1

准备插入3   因为队尾是1,1<3，删除1 ，插入3,3为队首也为队尾

准备插入-1  因为3>-1,单调递减，插入-1   这时候队列为 3 -1   前三个的最大值为队首3

准备插入-3，单调递减 ，3 -1 -3，第二个三个的和最大值也为队首，为3

准备插入5   单调递减，把-3 -1 3全部删除，插入5，为首元素，这时候最大值为5,

以此类推。

代码：

#include <iostream>
#include <stdio.h>
using namespace std;
const int maxn=1000005;
int n,k;
int q1[maxn],q2[maxn],num[maxn],Min[maxn],Max[maxn];
int front1,rear1,front2,rear2,cnt1,cnt2;

void in1(int i)//入队，单调递增，保存最小值
{
    while(front1<=rear1&&num[q1[rear1]]>num[i])
        rear1--;
    q1[++rear1]=i;
}
void in2(int i)//单调递减，保存最大值
{
    while(front2<=rear2&&num[q2[rear2]]<num[i])
        rear2--;
    q2[++rear2]=i;
}
void out1(int i)
{
    if(q1[front1]<=i-k)
        front1++;
    Min[cnt1++]=num[q1[front1]];
}
void out2(int i)
{
    if(q2[front2]<=i-k)
        front2++;
    Max[cnt2++]=num[q2[front2]];
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        front1=front2=cnt1=cnt2=0;
        rear1=rear2=-1;
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        for(int i=1;i<k;i++)//前k-1个数只入队，因为不可能达到出队条件
        {
            in1(i);in2(i);
        }
        for(int i=k;i<=n;i++)
        {
            in1(i);out1(i);
            in2(i);out2(i);
        }
        for(int i=0;i<cnt1-1;i++)
            printf("%d ",Min[i]);
        printf("%d\n",Min[cnt1-1]);
        for(int i=0;i<cnt2-1;i++)
            printf("%d ",Max[i]);
        printf("%d\n",Max[cnt2-1]);
    }
    return 0;
}