poj 2823 Sliding Window（RMQ水题）

题目链接：http://poj.org/problem?id=2823

Sliding Window
Time Limit: 12000MS Memory Limit: 65536K
Total Submissions: 81382 Accepted: 22902
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
Source

POJ Monthly–2006.04.28, Ikki

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分析：

注意提交的时候不要选g++,不然会超时
直接上代码：

// #include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;
const int N=1e6+5;

int arr[N];
int Min[N];
int Max[N];
int dp[N][2];
int dp2[N][2];

int init(int n,int tmp)
{
	for(int i=1;i<=n;i++)
	{
		dp[i][0]=arr[i-1];
		dp2[i][0]=arr[i-1];
	}
	int v=0;
	for(int j=1;j<=tmp;j++)
	{
		for(int i=1;i+(1<<j)-1<=n;i++)
		{
			dp[i][v^1]=min(dp[i][v],dp[i+(1<<j-1)][v]);
			dp2[i][v^1]=max(dp2[i][v],dp2[i+(1<<j-1)][v]);
		}
		v^=1;
	}
	return v;

}


int main()
{
	int n, k;
	scanf("%d%d",&n,&k);

	for(int i=0;i<n;i++)
		scanf("%d",&arr[i]);
	int tmp=1.0*log(1.0*k)/log(1.0*2);
	int v=init(n,tmp);
	int cnt=0;
	
	for(int i=1;i+k-1<=n;i++)
	{
		
		Min[++cnt]=min(dp[i][v],dp[i+k-1-(1<<tmp)+1][v]);
		Max[cnt]=max(dp2[i][v],dp2[i+k-1-(1<<tmp)+1][v]);
		
	}
	for(int i=1;i<=cnt;i++)
		printf("%d%c",Min[i],i==cnt?'\n':' ');
	for(int i=1;i<=cnt;i++)
		printf("%d%c",Max[i],i==cnt?'\n':' ');
		
	
	
	return 0;
}