ACM题目————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)

 

题目大概意思就是说：n头牛，两个方向，向后B，向前F，每次转K头牛，最少转cnt次，

最终使所有的牛向前。求最少的K以及对应的cnt。

 

代码如下：

#include <stdio.h>
#include <string.h>

#define maxn 5010

int cow[maxn], N, K, M; // 奶牛们的初始位置,0前1后
int tra[maxn]; // 是否翻转第i头牛,1翻

int cal(int k)   // 翻转顺序从左往右
{
    int ans = 0, i, sum = 0;
    for(i = 0; i + k <= N; ++i)
    {
        tra[i] = 0;
        if((sum + cow[i]) & 1)
        {
            tra[i] = 1;
            ++ans;
        }
        sum += tra[i]; // 尺取法
        if(i - k + 1 >= 0)
            sum -= tra[i-k+1];
    }
    // 检查剩下的是否有反向奶牛
    for( ; i < N; ++i)
        if((sum + cow[i]) & 1)
            return -1;
        else if(i - k + 1 >= 0)
            sum -= tra[i-k+1];
    return ans;
}

int main()
{
    char ch[2];
    int i, m, k;
    scanf("%d", &N);
        for(i = 0; i < N; ++i)
        {
            scanf("%s", ch);
            if(ch[0] == 'B') cow[i] = 1;
            else cow[i] = 0;
        }
        M = N;
        K = 1;
        for(k = 1; k <= N; ++k)
        {
            m = cal(k);
            if(m >= 0 && m < M)
            {
                M = m;
                K = k;
            }
        }
        printf("%d %d\n", K, M);
    return 0;
}