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POJ 2823 Sliding Window （單調隊列）

編輯：關於C++

Sliding Window Time Limit: 12000MS Memory Limit: 65536K Total Submissions: 42278 Accepted: 12479 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

題意：給一個序列，求長度為k的框中最大最小的，

思路：單調隊列，維護長度為k的框中的最大，然後再來一次單調隊列維護最小的

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __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 1000005

int 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;
}