[ACM] poj 2823 Sliding Window（单调队列）

时间：2014-05-02 22:56:12 收藏：0 阅读：570

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 36212		Accepted: 10723
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.

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：

被单调队列搞得头痛欲裂，这里不解释了。

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;
}

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