poj 3276 Face The Right Way(开关问题)

Face The Right Way

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4874		Accepted: 2271

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)

tips:基于这些事实

1.如果一个区间的第一头牛的方向不对的话，那么此区间必须翻转。

2.对同一区间的多次翻转是没有意义的，因此某一区间的翻转次数要么是0要么是1。利用f数组记录

3.记录下每一头牛的翻转次数，利用翻转次数和初始方向的关系判断次牛在区间初始位置时，是否需要翻转。

4.通过递推关系（前一头牛翻转次数）得到这头牛翻转的次数

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

int n,dir[5003],f[5003];//dir代表方向，0前，1后；f代表区间[i,i+x-1]是否翻转 

int cal(int x)//x代表区间大小 
{
	memset(f,0,sizeof(f));
	int ret=0;int sum=0;
	for(int i=1;i+x-1<=n;i++)
	{
		if((sum+dir[i])%2!=0)//如果本区间的第一头牛的方向朝后 
		{
			ret++;
			f[i]=1;
			sum+=f[i];
		}
	//	sum+=f[i];            //计算下一个区间的第一头牛被翻转的次数 
		if(i-(x-1)>=1)sum-=f[i-x+1]; //由递推公式获得的下一头牛翻转的次数 
	}
	
	//判断最后一个区间牛 地方向 。本区间的牛无法再继续翻转，方向必须为F 
	for(int i=n-x+2;i<=n;i++)
	{
		if((sum+dir[i])%2==1){
			return -1;
		} 
		if(i-(x-1)>=1)sum-=f[i-x+1]; 
	} 
	return ret;
	
	
}
int main()
{
	while(cin>>n)
	{
		for(int i=1;i<=n;i++){
			char x;cin>>x;
			dir[i]=(x=='F'?0:1);
	 
		}
		int M=n;int K=1;//初始化操作次数与
		for(int k=1;k<=n;k++)
		{
			int m=cal(k);
			if(m<M&&m>=0)//如果可行并且更优 
			{
				M=m;K=k; 
			}
		 } 
		 cout<<K<<" "<<M<<endl; 
	}
	
	
	return 0;
}