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：

? ? ? 一個長度為k的窗口，向右滑動，要求輸出每一次滑動，窗口內元素的最小值和最大值。（這樣的話每行輸出的元素就應該是n-k+1個元素咯，因為滑動的窗口從【1-1】先得增長到【1-k】吧）

解題報告：

? ? ? 雙端隊列，也可以用模擬雙端隊列做。

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

#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）

#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：

? ? ?怎么解決一下超時問題？

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總結

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