POJ 3276 Face The Right Way

传送门：http://poj.org/problem?id=3276

Face The Right Way

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4504		Accepted: 2086

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)；

这是翻转问题，

1. 如果第一个是方向相反，那么必须翻转，同理第二个第二头牛同上。

2.如果翻转次数为奇数，那么该牛需要翻转，否则不必要翻转。

3.在最后末尾的时候，即距离最后一个差k-1个的时候，就检测最后几个是否全部为正确的方向，如果最后k-1个不全为正确的方向，那么这种翻转方法不行，反之则行。

这是看书的解题思路的来的，深深的感到了智商的碾压。有点灰心。

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

const int MAX_N=6000;
int dir[MAX_N],f[MAX_N];
int N;

int calc(int k )
{
    int res=0;
    int sum=0;
    memset(f,0,sizeof(f));

    for(int i=0; i+k<=N; i++){
        if((sum+dir[i])%2!=0){
            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)%2!=0)
            return -1;
        if(i-k+1>=0){
            sum-=f[i-k+1];
        }

    }
    return res;
}

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
",K,M);
}

int main()
{
    cin>>N;
    for(int i=0; i<N; i++){
        char tmp;
        cin>>tmp;
        tmp=='F'?dir[i]=0:dir[i]=1;
    }
    solve();
    return 0;
}