| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 53086 | Accepted: 15227 | |
| 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
模拟滑动区间，用单调队列维护最值即可。本来正常做法是用单调队列存下标，然而蠢蠢的我又多开了俩数组……这样效率会降低，好在还是过了。
1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 #include<map>
5 using namespace std;
6 const int mxn=2000200;
7 int n,k;
8 int num;
9 int qmx[mxn],qmi[mxn];
10 int lmx[mxn],lmi[mxn];
11 int ansmx[mxn];
12 int main(){
13 scanf("%d%d",&n,&k);
14 int i,j;
15
16 int h1=1,t1=0;
17 int h2=1,t2=0;
18 for(int cnt=1;cnt<=n;cnt++){
19 scanf("%d",&num);
20 //max
21 while(h1<=t1 && num>=qmx[t1])t1--;
22 qmx[++t1]=num;
23 lmx[t1]=cnt;//下标
24
25
26 //min
27 while(h2<=t2 && num<=qmi[t2])t2--;
28 qmi[++t2]=num;
29 lmi[t2]=cnt;
30
31 if(lmx[t1]-lmx[h1]>=k)h1++;
32 if(lmi[t2]-lmi[h2]>=k)h2++;
33 // printf("test: h1:%d t1:%d\n",h1,t1);
34 ansmx[cnt]=qmx[h1];
35 if(cnt>=k)printf("%d ",qmi[h2]);
36 }
37 printf("\n");
38 for(i=k;i<=n;i++)printf("%d ",ansmx[i]);
39
40 return 0;
41 }