POJ3276 Face The Right Way

POJ3276 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 ≤ KN) 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

1
2
3
4
5
6
7
8
7
B
B
F
B
F
B
B

Sample Output

1
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)

Answer

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#include <iostream>
#include <cstring>
using namespace std;

int N, dir[5010], f[5010];
int calc(int K);

int main() {
char t;
cin >> N;
for (int i = 0; i < N; ++i) {
cin >> t;
dir[i] = t == 'B' ? 1 : 0;
}
int K = 1, M = N;
for (int k = 1; k <= N; ++k) {
int m = calc(k);
if (m >= 0 && M > m) {
M = m;
K = k;
}
}
cout << K << " " << M << endl;
return 0;
}

int calc(int K) {
memset(f, 0, sizeof(f));
int res = 0, sum = 0;
for (int i = 0; i + K - 1 < N; ++i) {
if((dir[i]+sum)&1) {
++res;
f[i] = 1;
}
sum += f[i];
if (i - K + 1>= 0) {
sum -= f[i-K+1];
}
}
for (int i = N-K+1; i < N; ++i) {
if((dir[i]+sum)&1) {
return -1;
}
if (i - K + 1 >= 0) {
sum -= f[i-K+1];
}
}
return res;
}

Analysis

  1. N 头牛或向前或向后,每次反转一个长为 K 的连续区间,求最小的反转次数及此时对应的最小的 K,使得牛全部向前。
  2. 从 1...N 遍历 K,每个 K 遍历一次牛的区间,没遇到一次向后则翻转一次。
  3. 为了减少翻转的操作复杂度,令 f 数组为记录是否反转,sum 为之前的反转次数(反转两次如同原样)。
# POJ

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