Sliding Window 鏈接：http://poj.org/problem?id=2823

Time Limit:?12000MS	?	Memory Limit:?65536K
?	?	?
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. 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

Source

POJ Monthly--2006.04.28, Ikki 題意：給定一個數列，從左至右輸出每個長度為m的數列段內的最小數和最大數。
數列長度：N<=106，m<=N #include<iostream> #include<cstdio> using namespace std;const int N = 1000005;int n,k,m,mn[N],mx[N],n1,n2;struct node{int id, data; };node dj[N],dz[N];//dz 遞增隊列 dj 遞減隊列 int main() {cin >> n >> k;int headj = 1,tail1 = 0,headz = 1,tail2 = 0;for(int i = 1; i <= n; i++){scanf("%d",&m);while(m <= dj[tail1].data && tail1 >= headj) tail1 --;//保持單調性 while(m >= dz[tail2].data && tail2 >= headz) tail2 --;dj[++tail1].data = m;dj[tail1].id = i;dz[++tail2].data = m;dz[tail2].id = i;if(i >= k){//開始記錄第n段中的最值 if(dj[headj].id <= i-k) mn[++n1] = dj[++headj].data;//過期了 else mn[++n1] = dj[headj].data;if(dz[headz].id <= i-k) mx[++n2] = dz[++headz].data;else mx[++n2] = dz[headz].data;}}for(int i = 1; i <= n1; i++)printf("%d ",mn[i]);cout<<endl;for(int i = 1; i <= n1; i++)printf("%d ",mx[i]);return 0; }

第一次手打隊列，以后還是用雙向吧

轉載于:https://www.cnblogs.com/EdSheeran/p/8406551.html

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