Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 28126 | Accepted: 8389 | |
| 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.
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
MYCode(线段树版本)
#include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAX 1000010 struct node { int lt; int rt; int mini; int most; }a[4*MAX]; int v[MAX]; int lt[MAX]; int rt[MAX]; int min(int a,int b){return a<b?a:b;} int max(int a,int b){return a>b?a:b;} void build(int s,int t,int step) { a[step].lt=s; a[step].rt=t; if(s==t) { a[step].mini=a[step].most=v[s]; return; } int mid=(a[step].lt+a[step].rt)/2; build(s,mid,2*step); build(mid+1,t,2*step+1); a[step].mini=min(a[2*step].mini,a[2*step+1].mini); a[step].most=max(a[2*step].most,a[2*step+1].most); } node query(int s,int t,int step) { if(s==a[step].lt&&t==a[step].rt) { return a[step]; } if(a[step].lt==a[step].rt)return a[step]; int mid=(a[step].lt+a[step].rt)/2; if(mid>=t) return query(s,t,2*step); else if(mid<s) return query(s,t,2*step+1); else { node tp1=query(s,mid,2*step); node tp2=query(mid+1,t,2*step+1); node tp; tp.lt=s; tp.rt=t; tp.mini=min(tp1.mini,tp2.mini); tp.most=max(tp1.most,tp2.most); return tp; } } int main() { int n,k; while(scanf("%d%d",&n,&k) != EOF) { int i; for(i=1;i<=n;i++) { scanf("%d",&v[i]); } build(1,n,1); int ct=0; for(i=1;i<=n-k+1;i++) { node ans=query(i,i+k-1,1); lt[ct]=ans.mini; rt[ct]=ans.most; ct++; } for(i=0;i<ct;i++) { printf("%d",lt[i]); if(i!=ct-1) printf(" "); } printf("\n"); for(i=0;i<ct;i++) { printf("%d",rt[i]); if(i!=ct-1) printf(" "); } printf("\n"); } }
MYCode:(单调队列版本)
#include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAX 1000010 int v[MAX]; int lt[MAX]; int rt[MAX]; int st[MAX]; int n; int k; void solve1() { int head=0,tail=0; int ct=0; int i; for(i=1;i<=n;i++) { while(head<tail&&v[st[tail-1]]<v[i])tail--; st[tail++]=i; while(i-st[head]>=k)head++; rt[ct++]=st[head]; } } void solve2() { int head=0,tail=0; int ct=0; int i; for(i=1;i<=n;i++) { while(head<tail&&v[st[tail-1]]>v[i])tail--; st[tail++]=i; while(i-st[head]>=k)head++; lt[ct++]=st[head]; } } int main() { while(scanf("%d%d",&n,&k) != EOF) { int i; for(i=1;i<=n;i++) { scanf("%d",&v[i]); } solve1(); solve2(); for(i=k-1;i<n;i++) { printf("%d",v[lt[i]]); if(i!=n-1); printf(" "); } printf("\n"); for(i=k-1;i<n;i++) { printf("%d",v[rt[i]]); if(i!=n-1); printf(" "); } printf("\n"); } }
| Run ID | User | Problem | Result | Memory | Time | Language | Code Length | Submit Time |
| 11039225 | ruffy | 2823 | Accepted | 43848K | 11094MS | C++ | 1959B | 2012-11-22 15:22:12 |
| 11039159 | ruffy | 2823 | Accepted | 11940K | 6735MS | C++ | 1208B | 2012-11-22 15:00:32 |
线段树11094MS
单调队列6735MS