【C++】Sliding Window

【來源】

POJ-2823
vjudge

【題目描述】

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.

【輸入格式】

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.

【輸出格式】

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.

【樣例輸入】

8 3
1 3 -1 -3 5 3 6 7

【樣例輸出】

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

【題目大意】

給您一個大小爲n的數組。有一個k大小的滑動窗口，它從數組的最左邊移動到最右邊。您只能在窗口中看到K數字。每次滑動窗口向右移動一個位置。給定一個序列，對於所有的i>=k，輸出序列中[i-k+1,i]內的最大值和最小值。

【解析】

單調隊列優化動態規劃模板題。
分別求min和max即可。

【代碼】

本代碼用C++提交。
若用G++提交請將O3優化去掉，並使用快讀。

#pragma GCC optimize(3,"Ofast","inline")
#pragma G++ optimize(3,"Ofast","inline")

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

#define RI                 register int
#define re(i,a,b)          for(RI i=a; i<=b; i++)
#define ms(i,a)            memset(a,i,sizeof(a))
#define MAX(a,b)           (((a)>(b)) ? (a):(b))
#define MIN(a,b)           (((a)<(b)) ? (a):(b))

using namespace std;

typedef long long LL;

const int N=1e6+5;
const int inf=1e9;

int n,k;
int a[N],q[N];

inline void calcmin() {
	int l=1,r=0;
	for(int i=1; i<=n; i++) {
		while(l<=r && a[i]<a[q[r]]) r--;
		q[++r]=i;
		while(l<=r && i-q[l]>=k) l++;
		if(i>=k) printf("%d ",a[q[l]]);
	}
	printf("\n");
}

inline void calcmax() {
	int l=1,r=0;
	for(int i=1; i<=n; i++) {
		while(l<=r && a[i]>a[q[r]]) r--;
		q[++r]=i;
		while(l<=r && i-q[l]>=k) l++;
		if(i>=k) printf("%d ",a[q[l]]);
	}
	printf("\n");
}

int main() {
	scanf("%d%d",&n,&k);
	for(int i=1; i<=n; i++) scanf("%d",&a[i]);
	calcmin();
	calcmax();
	return 0;
}