POJ 3276 Face The Right Way 开关问题

Face The Right Way

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4367		Accepted: 2025

Description

Farmer John has arranged his N (1 ≤ N ≤ 5,000) cows in a row and many of them are facing forward, like good cows. Some of them are facing backward, though, and he needs them all to face forward to make his life perfect.

Fortunately, FJ recently bought an automatic cow turning machine. Since he purchased the discount model, it must be irrevocably preset to turn K (1 ≤ K ≤ N) cows at once, and it can only turn cows that are all standing next to each other in line. Each time the machine is used, it reverses the facing direction of a contiguous group of K cows in the line (one cannot use it on fewer than K cows, e.g., at the either end of the line of cows). Each cow remains in the same*location* as before, but ends up facing the *opposite direction*. A cow that starts out facing forward will be turned backward by the machine and vice-versa.

Because FJ must pick a single, never-changing value of K, please help him determine the minimum value of K that minimizes the number of operations required by the machine to make all the cows face forward. Also determine M, the minimum number of machine operations required to get all the cows facing forward using that value of K.

Input

Line 1: A single integer: N 
Lines 2..N+1: Line i+1 contains a single character, F or B, indicating whether cow i is facing forward or backward.

Output

Line 1: Two space-separated integers: K and M

Sample Input

7
B
B
F
B
F
B
B

Sample Output

3 3

Hint

For K = 3, the machine must be operated three times: turn cows (1,2,3), (3,4,5), and finally (5,6,7)

Source

USACO 2007 March Gold

 

题目：http://poj.org/problem?id=3276

来自挑战程序设计竞赛

 

分析：首先这是个反转（开关问题），不难发现，每个区间反转两次及以上是没有意义的；其次区间反转的顺序是无所谓的。对于每一个k，我们从左起遍历每一个区间一次，每个区间是否反转取决于本区间第一个元素的状态。而每个元素的状态取决于它前k-1个元素的反转情况；设f[i] = 1代表反转第i个区间，则若sum(f[i-k+1] +...+f[i-1])为奇数，第i个元素此时是反转的，这个区间和可以在常量时间内计算出来，这样算法复杂度为O(n^2);