POJ 2823 Sliding Window (monotonous Queue)
Sliding Window
Time Limit:12000 MS |
|
Memory Limit:65536 K |
Total Submissions:42278 |
|
Accepted:12479 |
Case Time Limit:5000 MS |
Description
An array of size
N≤ 106 is given to you. There is a sliding window of size
KWhich is moving from the very left of the array to the very right. You can only see
KNumbers 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
KIs 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
NAnd
KWhich are the lengths of the array and the sliding window. There are
NIntegers 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 31 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 33 3 5 5 6 7
Source
POJ Monthly -- 2006.04.28, Ikki
For a sequence, find the maximum and minimum number in the box with the length of k,
Train of Thought: monotonous queue, maintain the maximum value in the box with a length of k, and then try to maintain the smallest monotonous queue.
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#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b) for(i = a; i < b; i++)#define free(i,a,b) for(i = a; i > =b;i--)#define mem(t, v) memset ((t) , v, sizeof(t))#define ssf(n) scanf("%s", n)#define sf(n) scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf printf#define bug pf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 1000005int a[N],que[N*2];int tail,head;int mi[N],ma[N];int le,ri;int n,k;inline void miinque(int i){ while(head
=a[que[tail-1]])tail--; que[tail++]=i;}inline void outque(int i){if(i-k>=que[head]) head++;}int main(){ while(~sff(n,k)) { int i; fre(i,1,n+1) sf(a[i]); head=tail=0; le=ri=0; fre(i,1,k) { miinque(i); } fre(i,k,n+1) { miinque(i); outque(i); mi[le++]=a[que[head]]; } head=tail=0; fre(i,1,k) { mainque(i); } fre(i,k,n+1) { mainque(i); outque(i); ma[ri++]=a[que[head]]; } fre(i,0,le) { if(i) pf(" "); pf("%d",mi[i]); } pf("\n"); fre(i,0,ri) { if(i) pf(" "); pf("%d",ma[i]); } pf("\n"); } return 0;}