POJ3276-Face The Right Way

Face The Right Way

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5064 Accepted: 2356

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

7BBFBFBB

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)

Source

USACO 2007 March Gold

题意：n头牛排成一列。每头牛或者向前或者向后。为了让所有牛都面向前方，每次可以将k头连续的牛转向1<=k<=n，求操作的最少次数M和对应的最小k
解题思路：枚举K,然后从左到右扫一遍，判断在该牛的前面第k头牛是否转向，若转向则总转向次数减一，然后判断该牛是否要转向

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;char ch[5009];int n,a[5009],visit[5009];int main(){while(~scanf("%d",&n)){for(int i=1;i<=n;i++){scanf("%s",ch+i);if(ch[i]=='F') a[i]=1;else a[i]=0;}int mi=INF,k;for(int i=1;i<=n;i++){int flag=1,sum=0,cnt=0;memset(visit,0,sizeof visit);for(int j=1;j<=n-i+1;j++){if(visit[j-i]&&j-i>=1) sum--;if((a[j]+sum)%2==0) visit[j]=1,sum++,cnt++;}for(int j=n-i+2;j<=n;j++){if(visit[j-i]&&j-i>=1) sum--;if((a[j]+sum)%2==0) flag=0;}if(flag&&mi>cnt) {k=i,mi=cnt;}}printf("%d %d\n",k,mi);}return 0;}