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.
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
正解：单调队列
解题报告：
单调队列裸题。发现提交用G++就会TLE，换成 C++就迷之AC了。。。
1 //It is made by jump~ 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <ctime> 9 #include <vector> 10 #include <queue> 11 #include <map> 12 #include <set> 13 #ifdef WIN32 14 #define OT "%I64d" 15 #else 16 #define OT "%lld" 17 #endif 18 using namespace std; 19 typedef long long LL; 20 const int MAXN = 2000011; 21 int n,k; 22 int ans[MAXN],ans2[MAXN]; 23 struct node{ 24 int val,id; 25 }a[MAXN]; 26 struct que{ 27 int head,tail; 28 node dui[MAXN]; 29 }a1,a2; 30 31 inline int getint() 32 { 33 int w=0,q=0; 34 char c=getchar(); 35 while((c<'0' || c>'9') && c!='-') c=getchar(); 36 if (c=='-') q=1, c=getchar(); 37 while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); 38 return q ? -w : w; 39 } 40 41 inline void work(){ 42 n=getint(); k=getint(); 43 for(int i=1;i<=n;i++) a[i].val=getint(),a[i].id=i; 44 a1.head=a2.head=1; 45 for(int i=1;i<=n;i++) { 46 if(i>k) { 47 if(i-a1.dui[a1.head].id>=k) a1.head++; 48 if(i-a2.dui[a2.head].id>=k) a2.head++; 49 } 50 while(a1.tail>=a1.head && a1.dui[a1.tail].val<=a[i].val) a1.tail--; 51 while(a2.tail>=a2.head && a2.dui[a2.tail].val>=a[i].val) a2.tail--; 52 a1.dui[++a1.tail]=a[i]; a2.dui[++a2.tail]=a[i]; 53 if(i>=k) ans[i]=a1.dui[a1.head].val,ans2[i]=a2.dui[a2.head].val; 54 } 55 for(int i=k;i<=n;i++) printf("%d ",ans2[i]); printf(" "); 56 for(int i=k;i<=n;i++) printf("%d ",ans[i]); 57 } 58 59 int main() 60 { 61 work(); 62 return 0; 63 }