TOJ 1701: Face The Right Way

1701: Face The Right Way

Time Limit(Common/Java):2000MS/20000MS     Memory Limit:65536KByte
Total Submit: 24            Accepted:9

Description

Farmer John has arranged hisN (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 turnK (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 ofK 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 ofK that minimizes the number of operations required by the machine to make all the cows face forward. Also determineM, the minimum number of machine operations required to get all the cows facing forward using that value ofK.

Input

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

Output

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

Sample Input

7BBFBFBB

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 March 2007

分析：枚举所有的K，求出对应的m，然后取最小的就行了。时间复杂度取决于求m的复杂度。对每一个k，从第1个位置开始，翻转所有反向的位置，然后用一个队列q记录对当前位置有影响的翻转（比如在位置1翻转了，那么直到位置k都会有影响），队列q的长度就是当前位置被翻转的次数，这样可以在O(n)时间内求出每一个m。总时间复杂度为O(k * n) = O(n^2)。