poj 2823 Sliding Window 單調隊列

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.

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

用一個長度爲k的窗在整數數列上移動，求窗裏面所包含的數的最小值最大值。

單調隊列定義：

a)從隊頭到隊尾，元素在我們所關注的指標下是遞減的（嚴格遞減，而不是非遞增），比如查詢如果每次問的是窗口內的最小值，那麼隊列中元素從左至右就應該遞增，如果每次問的是窗口內的最大值，則應該遞減，依此類推。這是爲了保證每次查詢只需要取隊頭元素。

b)從隊頭到隊尾，元素對應的時刻（此題中是該元素在數列a中的下標）是遞增的，但不要求連續，這是爲了保證最左面的元素總是最先過期，且每當有新元素來臨的時候一定是插入隊尾。

代碼：

#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;
}