Sliding Window
链接：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.
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
题意：给定一个数列，从左至右输出每个长度为m的数列段内的最小数和最大数。
数列长度：N<=106，m<=N
数列长度：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 递增队列 dj 递减队列 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 --;//保持单调性 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){ //开始记录第n段中的最值 if(dj[headj].id <= i-k) mn[++n1] = dj[++headj].data;//过期了 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; }
第一次手打队列，以后还是用双向吧
