【POJ 3276 Face The Right Way】+ 尺取法 + 开关问题

Face The Right Way
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 4439 Accepted: 2062

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)

枚举 K 的值~尺取法推进首位置~若需要反转~该位置标记~需要反转次数加一~最后特判下最后K个位置是否符合题意~更新最优解~~

AC代码：

#include
#include
const int MAXN = 5010;
int d[MAXN],f[MAXN],n;
char c[2];
int calc(int k){
   memset(f,0,sizeof(f));
   int ans = 0 , cut = 0;
   for(int i = 0 ; i + k <= n ; i++){
    if((d[i] + ans) % 2 != 0){ // 在该区间内的反转总次数 + 当前是否需要反转
        cut++; f[i] = 1; // 改区间反转过加一
    }
    ans += f[i];
    if(i - k + 1 >= 0) // 推进首位置
        ans -= f[i - k + 1];
   }
   for(int i = n - k + 1 ; i < n ; i++){
    if((d[i] + ans) % 2 != 0) return -1;
    if(i - k + 1 >= 0) ans -= f[i - k + 1];
   }
   return cut;
}
void solve(){
    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;
        }
    }
    printf("%d %d\n",K,M);
}
int main()
{
    scanf("%d",&n);
    for(int i = 0 ; i < n ; i++){
        scanf("%s",c);
        d[i] = c[0] == 'B' ? 1 : 0;
    }
    solve();
    return 0;
}