题目
有一个长为 nn 的序列 aa,以及一个大小为 kk 的窗口。现在这个从左边开始向右滑动,每次滑动一个单位,求出每次滑动后窗口中的最大值和最小值。
思路
维护一个单调队列即可。
代码实现
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=0;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005
#define fi first
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
char ch=getchar(); int x=0, f=1;
while(ch<'0'||ch>'9') {
if(ch=='-') f = -1;
ch=getchar();
}
while('0'<=ch&&ch<='9') {
x=x*10+ch-'0';
ch=getchar();
} return x*f;
}
struct node {
static const int maxn=1000001;
int n,k,a[maxn];
int p[maxn],q[maxn];
int head,tail;
void rea () {
scanf ("%d %d",&n,&k);
rep (i,1,n) scanf ("%d",&a[i]);
}
void queue_max () {
head=1;
tail=0;
rep (i,1,n) {
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]);
}
cout<<endl;
}
void queue_min () {
head=1;
tail=0;
rep (i,1,n) {
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 ("
");
}
}solve;
int main () {
// freopen ("data.in","r",stdin);
solve.rea ();
solve.queue_min();
solve.queue_max ();
// fclose (stdin);
return 0;
}