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POJ3276:Face The Right Way——开关问题、简单DP

牛头人约翰和牛的故事。

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)


题目大意是有N头牛排成一列,所有牛头随机超前或者朝后,现在需要让所有牛头都朝前方。要确定一个连续翻转的区间K满足条件,并且使翻转次数最小。

分析:
如果按照枚举做肯定会TLE,首先排除。

交换区间翻转的顺序对结果是没有影响的。而且没有必要对同一个区间进行两次以上的翻转。因此问题转化成了求需要被翻转区间的集合。
因此首先从最左侧牛开始考虑,如果该牛朝向前方,则不需要翻转这个区间;反之如果该牛朝后,则必须要处理这个区间,并且再次之后不需要考虑此最左的区间,区间范围-1。循环往返,即可求出最少翻转次数。

参考代码:


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