POJ 2823Sliding Window(單調佇列水題)

果7發表於2013-11-03
佇列
Sliding Window
Time Limit: 12000MS   Memory Limit: 65536K
Total Submissions: 33362   Accepted: 9918
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



題目意思就不多說了，直接用單調佇列，自己沒寫輸入優化時間還少一點，不過都需要好5s+。單調佇列模板題目，純屬娛樂。。預祝明天8086能在南京現場賽上取得好成績，fighting!!!!

題目地址：Sliding Window


AC程式碼：
#include<iostream>
#include<cstdio>
using namespace std;

struct node
{
    int num;
    int i;
}q1[1000006],q2[1000006];

int res1[1000006],res2[1000006];

int getint()
{
    int ans=0;
    int flag=0;
    char ch=getchar();
    while(ch<'0'||ch>'9'||ch=='-')
    {
        if(ch=='-') flag=1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        ans=ans*10+(ch-'0');
        ch=getchar();
    }
    if(flag) ans=-ans;
    return ans;
}

int main()
{
    int n,k,i;
    while(cin>>n>>k)
    {
        int top1,top2,tail1,tail2;
        top1=top2=tail1=tail2=0;
        int p;
        for(i=0;i<n;i++)
        {
            //p=getint();
            scanf("%d",&p);
            while(top1<tail1&&q1[tail1-1].num<p) tail1--;
            q1[tail1].i=i,q1[tail1++].num=p;
            while(i-q1[top1].i>=k) top1++;
            while(top2<tail2&&q2[tail2-1].num>p) tail2--;
            q2[tail2].i=i,q2[tail2++].num=p;
            while(i-q2[top2].i>=k) top2++;
            if(i>=k-1) res1[i-k+1]=q1[top1].num,res2[i-k+1]=q2[top2].num;
        }

        printf("%d",res2[0]);
        for(i=1;i<=n-k;i++)
            printf(" %d",res2[i]);
        printf("\n%d",res1[0]);
        for(i=1;i<=n-k;i++)
            printf(" %d",res1[i]);
        printf("\n");
    }
    return 0;
}

/*
8 3
-1 2 3 1 3 5 6 7
8 3
1 3 -1 -3 5 3 6 7
*/

//5344MS  未讀入優化
//7079MS  優化輸入之後。。。


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