poj 2823 Sliding Window（单调队列）
题目链接
Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 39725 | Accepted: 11747 | |
| Case Time Limit: 5000MS | ||
Description
An array of size
n ≤ 10
6 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.
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
题意：数n个数的数列a 。求所有长度为K的区间的最大值和最小值。
题解：单调队列的基本运用。我们让区间，最开始为a数列前k个数，每次后移一位。先谈最大值的情况，由于区间不断后移，若当前区间中， i>j 且 ai > aj ，那么aj 一定不能成为后面的区间的最优解。利用这个性质，我们可以维护一个单调的双端队列，队列里面存可能成为最优解的数。
队列里面的数要满足两个单调性，元素的数值按从大到小排列，元素在a数列中的下标按从小到大排列。这样队首元素为当前区间的最优值。
插入操作：插入的数下标一定比队尾大。假设插入的数大于队尾，那么插入的数一定比队尾更优，删除队尾，直到队尾元素大于插入的数或队列为空，将数插入队列。在插入操作中保证队列的两个单调性。
删除操作：随着区间的移动，我们要将队列里面不在区间里的数删除。由于下标按最小到大排列的，所以我们从队首开始删除，直至队首满足条件。
求最小值同理。
由于每个数插入、删除的复杂度为O（1），只入队和出队一次，所以总复杂度O（n）。
代码如下：
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<iostream>
#include<string.h>
#include<string>
#include<math.h>
#include<stdlib.h>
#define inff 0x3fffffff
#define eps 1e-8
#define nn 1100000
#define mod 1000000007
typedef long long LL;
const LL inf64=LL(inff)*inff;
using namespace std;
int n,k;
int a[nn];
int ans1[nn],ans2[nn];
int que1[nn],que2[nn];
int l1,r1,l2,r2;
void add1(int id)
{
while(r1>l1)
{
if(a[que1[r1-1]]<=a[id])
{
r1--;
}
else
{
que1[r1++]=id;
return ;
}
}
que1[r1++]=id;
}
void dele1(int id)
{
while(l1<r1)
{
if(que1[l1]<=id)
l1++;
else
break;
}
}
void add2(int id)
{
while(l2<r2)
{
if(a[que2[r2-1]]>=a[id])
{
r2--;
}
else
{
que2[r2++]=id;
return ;
}
}
que2[r2++]=id;
}
void dele2(int id)
{
while(l2<r2)
{
if(que2[l2]<=id)
l2++;
else
break;
}
}
int main()
{
int i;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
l1=r1=l2=r2=0;
int cnt=0;
for(i=1;i<=k;i++)
{
add1(i);
add2(i);
}
ans1[++cnt]=que1[l1];
ans2[cnt]=que2[l2];
for(i=k+1;i<=n;i++)
{
add1(i);
add2(i);
dele1(i-k);
dele2(i-k);
ans1[++cnt]=que1[l1];
ans2[cnt]=que2[l2];
}
for(i=1;i<=cnt;i++)
{
printf("%d%c",a[ans2[i]],i==cnt?'\n':' ');
}
for(i=1;i<=cnt;i++)
{
printf("%d%c",a[ans1[i]],i==cnt?'\n':' ');
}
}
return 0;
}