POJ 2823 Sliding Window（单调队列）

Sliding Window

Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 61914 Accepted: 17733Case 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 positionMinimum valueMaximum value[1  3  -1] -3  5  3  6  7 -13 1 [3  -1  -3] 5  3  6  7 -33 1  3 [-1  -3  5] 3  6  7 -35 1  3  -1 [-3  5  3] 6  7 -35 1  3  -1  -3 [5  3  6] 7 36 1  3  -1  -3  5 [3  6  7]37

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 31 3 -1 -3 5 3 6 7

Sample Output

-1 -3 -3 -3 3 33 3 5 5 6 7

Source

POJ Monthly--2006.04.28, Ikki

题意：

滑动窗口，一次只能看k个数字，求每次窗口中的最大值最小值。

POINT：

单调队列。

#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <algorithm>using namespace std;#define  LL long longconst int N = 1000000+10;struct node{    int x,y;}v[N];int a[N],n,k;int ans[N];void getmax(){    int cnt=0;    int head=1,end=0;    for(int i=1;i<k;i++)    {        while(head<=end&&v[end].x<=a[i]) end--;        v[++end].x=a[i],v[end].y=i;    }    for(int i=k;i<=n;i++)    {        while(head<=end&&v[end].x<=a[i]) end--;        v[++end].x=a[i],v[end].y=i;        while(i-v[head].y>=k) head++;        ans[++cnt]=v[head].x;    }    for(int i=1;i<=cnt;i++)    {        if(i-1) printf(" ");        printf("%d",ans[i]);    }    printf("\n");}void getmin(){    int cnt=0;    int head=1,end=0;    for(int i=1;i<k;i++)    {        while(head<=end&&v[end].x>=a[i]) end--;        v[++end].x=a[i],v[end].y=i;    }    for(int i=k;i<=n;i++)    {        while(head<=end&&v[end].x>=a[i]) end--;        v[++end].x=a[i],v[end].y=i;        while(i-v[head].y>=k) head++;        ans[++cnt]=v[head].x;    }    for(int i=1;i<=cnt;i++)    {        if(i-1) printf(" ");        printf("%d",ans[i]);    }    printf("\n");    }int main(){    scanf("%d %d",&n,&k);            for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        getmin();        getmax();    }