Sliding window (Luogu 1886)

Title Description

Now there are a bunch of numbers with N numbers (n<=10^6), and a window of size K. Now this swipe from the left to the right, sliding one unit at a time to find the maximum and minimum values in the window after each slide.

For example:

The array is [1 3-1-3 5 3 6 7], and k = 3.

input/output format

Input format:

Enter a total of two lines, the first behavior n,k.

The second behavior n number (<int_max).

Output format:

Output a total of two lines, the first behavior of each window sliding minimum value

Second behavior the maximum value of each window slide

input/Output sampleInput Sample

8 31 3-1-3 5 3 6 7

Output Sample

-1-3-3-3 3 33 3 5 5 6 7

Description

50% of data, n<=10^5

100% of data, n<=10^6

Ideas:

A template, thought very well understand, I think difficult in judging interval, super you have to head++, not monotonous you have to tail--

Code

#include <stdio.h>#include<algorithm>using namespacestd;Const intmx=1e6+1;intN,K,P[MX],Q[MX],A[MX];voidGetmax () {intHead=1, tail=0;  for(intI=1; i<=n;++i) { while(Head<=tail && q[tail]<=a[i]) tail--; q[++tail]=A[i]; P[tail]=i;  while(p[head]<=i-k) head++; if(i>=k) printf ("%d", Q[head]); }}voidgetmin () {intHead=1, tail=0;  for(intI=1; i<=n;++i) { while(Head<=tail && q[tail]>=a[i]) tail--; q[++tail]=A[i]; P[tail]=i;  while(p[head]<=i-k) head++; if(i>=k) printf ("%d", Q[head]); } printf ("\ n");}intMain () {scanf ("%d%d",&n,&k);  for(intI=1; i<=n;++i) {scanf ("%d",&A[i]);    } getmin ();    Getmax (); return 0;}/*8 to 3-1-3 5 3 6 7*/

Sliding window (Luogu 1886)