poj2823 单调队列

刚刚学习了单调队列，来练习一道，等会去试试多重背包。



如题：http://poj.org/problem?id=2823

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 47025		Accepted: 13585
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


思路:单调队列也就是在一个队列中维护一个单调不减或单调不增的序列，每次用的时候从头取出来。插入的时候在尾部，并在头部删除无效的值，具体百度。



#include<iostream>

#include<cstdio>

#include<cstring>

#include<vector>

using namespace std;

int a[1000005];

typedef pair<int,int>P;

int top1,rear1,top2,rear2;

P que1[1000005]; //单调减

P que2[1000005]; //单调加

void insert(int x,int i)

{

while(rear1>top1&&x<que1[rear1-1].first)

rear1--;

que1[rear1++]=P(x,i);

while(rear2>top2&&x>que2[rear2-1].first)

rear2--;

que2[rear2++]=P(x,i);

}

int main()

{

// freopen("C:\\1.txt","r",stdin);

int n,k;

cin>>n>>k;

int i;

for(i=1;i<=n;i++)

cin>>a[i];

top1=rear1=0;

top2=rear2=0;

vector<int>res1;

vector<int>res2;

for(i=1;i<=k;i++)

{

insert(a[i],i);

}

for(i=k+1;i<=n;i++)

{

res1.push_back(que1[top1].first);

res2.push_back(que2[top2].first);

insert(a[i],i);

while(que1[top1].second<i-k+1)

top1++;

while(que2[top2].second<i-k+1)

top2++;

}

res1.push_back(que1[top1].first);

res2.push_back(que2[top2].first);

for(i=0;i<res1.size();i++)

{

printf("%d%c",res1[i],i==res1.size()-1?'\n':' ');

}

for(i=0;i<res2.size();i++)

{

printf("%d%c",res2[i],i==res2.size()-1?'\n':' ');

}

return 0;

}