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POJ 2823 单调队列

2024-07-04 05:58:44   230人阅读

Sliding Window
Time Limit: 12000MS Memory Limit: 65536K
Total Submissions: 36469 Accepted: 10803
Case Time Limit: 5000MS

Description

An array of size n ≤ 106 is given to you. There is a sliding window of sizek which is moving from the very left of the array to the very right. You can only see thek 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 integersn and k which are the lengths of the array and the sliding window. There aren 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

Source


求每一个长度为k的区间的最大值，最小值，把它们都输出来。

与上面的那道很类似，用两个单调队列，一个递增，一个递减，然后从左向右扫描一遍即可。

POJ G++ tle，C++勉强ac，卡时较严重。

代码：

/* ***********************************************
Author :_rabbit
Created Time :2014/5/12 23:07:15
File Name :20.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int a[1001000],que1[1001000],que2[1001000],ans1[1001000],ans2[1000100];
int main()
{
     //freopen("data.in","r",stdin);
     //freopen("data.out","w",stdout);
     int n,k;
	 while(~scanf("%d%d",&n,&k)){
		 for(int i=1;i<=n;i++)scanf("%d",&a[i]);
		 int head=1,end=k,front1=0,tail1=0,front2=0,tail2=0;
		 for(int i=1;i<=k;i++){
			 while(front1<tail1&&a[que1[tail1-1]]>a[i])tail1--;//递增序列
			 que1[tail1++]=i;
			 while(front2<tail2&&a[que2[tail2-1]]<a[i])tail2--;//递减序列
			 que2[tail2++]=i;
		 }
		 ans1[1]=a[que1[front1]];ans2[1]=a[que2[front2]];
		 for(int i=k+1;i<=n;i++){
			 head=i-k+1;end=i;
			 while(front1<tail1&&a[que1[tail1-1]]>a[i])tail1--;
			 que1[tail1++]=i;
			 while(que1[front1]<head)front1++;
			 ans1[i-k+1]=a[que1[front1]];
			 while(front2<tail2&&a[que2[tail2-1]]<a[i])tail2--;
			 que2[tail2++]=i;
			 while(que2[front2]<head)front2++;
			 ans2[i-k+1]=a[que2[front2]];
		 }
		 for(int i=1;i<=n-k+1;i++)printf("%d%c",ans1[i],i==n-k+1?‘\n‘:‘ ‘);
		 for(int i=1;i<=n-k+1;i++)printf("%d%c",ans2[i],i==n-k+1?‘\n‘:‘ ‘);
	 }
     return 0;
}



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