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Face The Right Way
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)
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题目大意：有N头奶牛有些奶牛面向前方，有些奶牛面向后方。现在有一台奶牛翻转机，可以一次翻转连续的K头奶牛（K未给出），求最少的翻转次数M，以及在M最小的前提下最小的K。
这题的思路是显然的，从左到右枚举，如果某头牛是朝向后面的，就把从它开始的k头牛翻转。关键在于如何优化翻转这个过程。
暴力翻转的话复杂度为$\mathcal{O}\left( kn^2\right)$，会TLE。
因为是区间操作，想到用线段树来维护，复杂度为$\mathcal{O}\left( kn\log_2 n\right)$，还是会TLE。
看了题解后才知道了一种神奇的方法——记录每个点与前面的点的关系（相同或不同），这样在翻转时就只需要$\mathcal{O}\left( 1\right)$的时间，总时间复杂度为$\mathcal{O}\left( kn\right)$，能够过了。
代码在此。
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