poj 2823 Sliding Window

poj 2823 Sliding Window
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

【分析】
单调队列模板
分别用两个单调队列对区间最大值和最小值进行维护
棒棒哒

【代码】

//poj 2823 Sliding Window//单调队列模板题 #include<iostream>#include<cstdio>#include<vector>#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;const int T=1000001;int n,k,t;int a[T],q[T];   //q[i]:队尾元素在a数组中的下标 int main(){    int i,j,x,y,h,t;    scanf("%d%d",&n,&k);    fo(i,1,n) scanf("%d",&a[i]);    h=1,t=0;    fo(i,1,n)    {        while(h<=t && a[q[t]]>=a[i]) t--;        q[++t]=i;        while(h<=t && q[t]-q[h]>=k) h++;        if(i>=k) printf("%d ",a[q[h]]);    }    printf("\n");    h=1,t=0;    fo(i,1,n)    {        while(h<=t && a[q[t]]<=a[i]) t--;        q[++t]=i;        while(h<=t && q[t]-q[h]>=k) h++;        if(i>=k) printf("%d ",a[q[h]]);    }    printf("\n");    return 0;}