【poj 2823】 Sliding Window

最新推荐文章于 2021-01-25 20:02:52 发布

TLECODE

最新推荐文章于 2021-01-25 20:02:52 发布

阅读量240

点赞数

分类专栏： poj 单调队列/单调栈

本文链接：https://blog.youkuaiyun.com/TLECOCE/article/details/78231269

版权

poj 同时被 2 个专栏收录

38 篇文章

订阅专栏

单调队列/单调栈

6 篇文章

订阅专栏

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

这道题可以用两个单调队列分别维护最大值和最小值，时间复杂度为O(n)(注意不要使用STL的queue，不然可能会TLE)，下面是程序：

#include<queue>
#include<stdio.h>
#include<iostream>
using namespace std;
const int N=1000005;
struct node{
	int w,t;
}a[N];
struct que{
	int q[N],l,r;
	que(){
		l=0;
		r=1;
	}
	void pop_front(){
		l=(l+1)%N;
	}
	void pop_back(){
		r=(r+N-1)%N;
	}
	void push_front(int n){
		q[l]=n;
		l=(l+N-1)%N;
	}
	void push_back(int n){
		q[r]=n;
		r=(r+1)%N;
	}
	int front(){
		return q[(l+1)%N];
	}
	int back(){
		return q[(r+N-1)%N];
	}
	bool empty(){
		return (l+1)%N==r;
	}
};
que q1,q2;
int ans[N][2];
int read(){
	int s=0,f=1;
	char c=getchar();
	while((c<'0'||c>'9')&&c!='-'){
		c=getchar();
	}
	if(c=='-'){
		f=-1;
	}
	while(c>='0'&&c<='9'){
		s*=10;
		s+=c-'0';
		c=getchar();
	}
	return s*f;
}
int main(){
	int n,k,i,tp;
	n=read();
	k=read();
	for(i=1;i<=n;i++){
		scanf("%d",&a[i].w);
		a[i].t=i;
	}
	for(i=1;i<=k;i++){
		tp=q1.back();
		while(!q1.empty()&&a[i].w<a[tp].w){
			q1.pop_back();
			tp=q1.back();
		}
		q1.push_back(i);
		tp=q2.back();
		while(!q2.empty()&&a[i].w>a[tp].w){
			q2.pop_back();
			tp=q2.back();
		}
		q2.push_back(i);
	}
	ans[0][0]=q1.front();
	ans[0][1]=q2.front();
	for(i=k+1;i<=n;i++){
		tp=q1.front();
		while(!q1.empty()&&a[tp].t<=i-k){
			q1.pop_front();
			tp=q1.front();
		}
		tp=q1.back();
		while(!q1.empty()&&a[i].w<a[tp].w){
			q1.pop_back();
			tp=q1.back();
		}
		q1.push_back(i);
		tp=q2.front();
		while(!q2.empty()&&a[tp].t<=i-k){
			q2.pop_front();
			tp=q2.front();
		}
		tp=q2.back();
		while(!q2.empty()&&a[i].w>a[tp].w){
			q2.pop_back();
			tp=q2.back();
		}
		q2.push_back(i);
		ans[i-k][0]=q1.front();
		ans[i-k][1]=q2.front();
	}
	for(i=0;i<n-k+1;i++){
		printf("%d ",a[ans[i][0]].w);
	}
	putchar('\n');
	for(i=0;i<n-k+1;i++){
		printf("%d ",a[ans[i][1]].w);
	}
	putchar('\n');
	return 0;
}