題干：

An array of size?n?≤ 10?6?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 positionMinimum valueMaximum 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

題目大意：

? ? ? 一個(gè)長(zhǎng)度為k的窗口，向右滑動(dòng)，要求輸出每一次滑動(dòng)，窗口內(nèi)元素的最小值和最大值。（這樣的話每行輸出的元素就應(yīng)該是n-k+1個(gè)元素咯，因?yàn)榛瑒?dòng)的窗口從【1-1】先得增長(zhǎng)到【1-k】吧）

解題報(bào)告：

? ? ? 雙端隊(duì)列，也可以用模擬雙端隊(duì)列做。

ac代碼：（用G++編譯器提交 4829ms，用C++編譯器 超時(shí)，將其中的cout改成printf后，G++，C++編譯器均超時(shí)。。。）

#include<iostream> #include<cstdio> #include<deque> const int MAX=1000000 + 5 ; using namespace std; struct Node {int x;int index; } s[MAX]; int n,k; void Min_Max() {deque<Node>dq;for(int i=1; i<=k; i++) {while(!dq.empty() && s[i].x<dq.back().x) dq.pop_back();dq.push_back(s[i]);}printf("%d",dq.front().x);for(int i=k+1; i<=n; i++){while(!dq.empty() && s[i].x<dq.back().x) dq.pop_back();dq.push_back(s[i]);while(dq.back().index-dq.front().index>=k) dq.pop_front();printf(" %d",dq.front().x);}cout<<endl; } void Max_Min() {deque<Node>q;for(int i=1; i<=k; i++) {while(!q.empty() && s[i].x>q.back().x) q.pop_back();q.push_back(s[i]);}printf("%d",q.front().x);for(int i=k+1; i<=n; i++) {while(!q.empty() && s[i].x>q.back().x) q.pop_back();q.push_back(s[i]);while(q.back().index-q.front().index>=k) q.pop_front();printf(" %d",q.front().x);} } int main() {scanf("%d %d",&n,&k);for(int i=1; i<=n; i++) {scanf("%d",&s[i].x);s[i].index=i;}Min_Max();Max_Min(); }

AC代碼：（C++編譯器9266ms，G++編譯器超時(shí)）

#include <iostream> #include <cstdio> #include <queue> #include <deque>using namespace std; typedef pair<int, int> P; #define maxn 1000000 + 10deque<P> Q1; deque<P> Q2; int n, k; int Min[maxn], Max[maxn];int main() {while(~scanf("%d%d", &n, &k)){while(!Q1.empty()) Q1.pop_back();while(!Q2.empty()) Q2.pop_back();int x;for(int i=1; i<=n; i++){scanf("%d", &x);while(!Q1.empty() && Q1.back().first >= x) Q1.pop_back();Q1.push_back(P(x, i));if(i >= k){while(!Q1.empty() && Q1.front().second <= i-k) Q1.pop_front();Min[i] = Q1.front().first;}while(!Q2.empty() && Q2.back().first <= x) Q2.pop_back();Q2.push_back(P(x, i));if(i >= k){while(!Q2.empty() && Q2.front().second <= i-k) Q2.pop_front();Max[i] = Q2.front().first;}}for(int i=k; i<=n; i++)i == n ? printf("%d\n", Min[i]) : printf("%d ", Min[i]);for(int i=k; i<=n; i++)i == n ? printf("%d\n", Max[i]) : printf("%d ", Max[i]);}return 0; }

總結(jié)：

? ? ?怎么解決一下超時(shí)問(wèn)題？

?

?

總結(jié)

以上是生活随笔為你收集整理的【POJ - 2823】 Sliding Window（单调队列 用双端队列实现或模拟队列）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。