2011-06-10 21:04:33Morris
a146. Sliding Window
http://zerojudge.tw/ShowProblem?problemid=a146
內容 ：
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.
輸入說明
：
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.
輸出說明
：
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.
範例輸入 ：
8 3 1 3 -1 -3 5 3 6 7 7 3 10 -26 89 80 27 84 73
範例輸出 ：
-1 -3 -3 -3 3 3 3 3 5 5 6 7 -26 -26 27 27 27 89 89 89 84 84
提示
：
// 輸入改成多筆測資
出處
：
POJ 2823 | ST(X) | Heap(X) | 單調隊列
(管理：morris1028)
作法 : 單調隊列
以後可以用來優化DP，在此先學著點
PS.
在此題效率是O(N)，常數是 2
比起其他tree or heap 的 O(Nlogk)，還會差好幾倍呢
作法 : 單調隊列
以後可以用來優化DP，在此先學著點
PS.
在此題效率是O(N)，常數是 2
比起其他tree or heap 的 O(Nlogk)，還會差好幾倍呢
/**********************************************************************************/
/* Problem: a146 "Sliding Window" from POJ 2823 | ST(X) | Heap(X) | 單調隊列*/
/* Language: C */
/* Result: AC (752ms, 2860KB) on ZeroJudge */
/* Author: morris1028 at 2011-06-08 23:34:38 */
/**********************************************************************************/
#include<stdio.h>
int A[1000001], Q[1000001], L[1000001];
int n, k;
void pushmin() {
int a, head = 1, tail = 0, t;
for(a = 1; a < k; a++) {
while(head <= tail && Q[tail] > A[a])
tail --;
Q[++tail] = A[a], L[tail] = a;
}
for(a = k; a <= n; a++) {
while(head <= tail && Q[tail] > A[a])
tail --;
Q[++tail] = A[a], L[tail] = a;
t = a - k;
while(L[head] <= t) head ++;
printf("%d%c", Q[head], (a < n) ? ' ' : '\n');
}
}
void pushmax() {
int a, head = 1, tail = 0, t;
for(a = 1; a < k; a++) {
while(head <= tail && Q[tail] < A[a])
tail --;
Q[++tail] = A[a], L[tail] = a;
}
for(a = k; a <= n; a++) {
while(head <= tail && Q[tail] < A[a])
tail --;
Q[++tail] = A[a], L[tail] = a;
t = a - k;
while(L[head] <= t) head ++;
printf("%d%c", Q[head], (a < n) ? ' ' : '\n');
}
}
int Input() {
char cha, flag = 1;
unsigned int x = 0;
while(cha = getchar())
if(cha != ' ' && cha != '\n' || cha == EOF) break;
if(cha != '-') x = x*10 + cha -48;
else flag = -1;
while(cha = getchar()) {
if(cha == ' ' || cha == '\n') break;
x = x*10 + cha-48;
}
return x * flag;
}
main() {
int a;
while(scanf("%d %d", &n, &k) == 2) {
if(k > n) k = n;
for(a = 1; a <= n; a++)
A[a] = Input();
pushmin();
pushmax();
}
return 0;
}