POJ 2823 Sliding Window(单调队列)

A - Sliding Window

Time Limit:12000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 2823

Description

An array of size  n ≤ 10  ⁶ 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  nintegers 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

题意:有一个区间，维护这个区间的最大值和最小值即可。裸的单调队列，开两个单调队列，分别维护最大值和最小值。没了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 100000000
#define maxn 2000000
using namespace std;
int num[maxn];
struct WTF {
	int k, w;
}q[maxn];
int main()
{
	int n, k, end, i,first,flag;
	WTF now;
	scanf("%d%d", &n, &k);
	
	for (i = 1;i <= n;i++)
	{
		scanf("%d", &num[i]);
		
	}
	///
	q[0].k = -INF;
	q[0].w = 0;
	end = 0;
	first = 1;
	for (i = 1;i < k;i++)
	{

		while (num[i] < q[end].k)
		{
			end--;
		}
		now.w = i;
		now.k = num[i];
		q[++end] = now;
	}
	flag = 0;
	for (;i <= n;i++)
	{
		while (num[i] < q[end].k&&end>=first)
		{
			end--;
		}
		now.w = i;
		now.k = num[i];
		q[++end] = now;
		flag++;
		while(q[first].w < flag)
			first++;
		printf("%d", q[first].k);
		if (i == n)
			printf("\n");
		else printf(" ");
	}
	///
	memset(q, 0, sizeof(q));
	q[0].k = INF;
	q[0].w = 0;
	end = 0;
	first = 1;
	for (i = 1;i < k;i++)
	{
		
		while (num[i] > q[end].k)
		{
			end--;
		}
		now.w = i;
		now.k = num[i];
		q[++end] = now;
	}
	 flag = 0;
	for (;i <= n;i++)
	{
		while (num[i] > q[end].k&&first<=end)
		{
			end--;
		}
		now.w = i;
		now.k = num[i];
		q[++end] = now;
		flag++;
		while (q[first].w < flag)
			first++;
		printf("%d", q[first].k);
		if (i == n)
			printf("\n");
		else printf(" ");
	}
	
}