POJ 3276 Face The Right Way

最新推荐文章于 2019-10-22 17:17:33 发布

原创最新推荐文章于 2019-10-22 17:17:33 发布 · 82 阅读

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本文探讨了一种使用自动牛群转向机将所有牛统一朝向的最优化算法。通过设定不同的连续转向牛只数量K，寻找最小化操作次数的方法。输入牛群初始朝向，输出最佳K值及最少操作步骤。

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Face The Right Way

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:6164		Accepted: 2854

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 Kcows, 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)

Source

USACO 2007 March Gold

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <set>
#include <map>

using namespace std;

typedef long long ll;

const int M=5000+5;

int v[M];
bool f[M];


int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        getchar();
        char c;
        scanf("%c",&c);
        if(c=='F'){
            v[i]=1;
        }
        else{
            v[i]=0;
        }
    }


    int res=n+5;
    int resx;

    for(int i=1;i<=n;i++){
        memset(f,0,sizeof(f));
        int ans=0;
        int sum=0;
        for(int j=1;j+i-1<=n;j++){
            if((sum+v[j])%2==0){
                f[j]=1;
                ans++;
            }
            sum+=f[j];
            if(j>=i){
                sum-=f[j-i+1];
            }
        }
        bool flag=1;
        for(int j=n-i+2;j<=n;j++){
            if((sum+v[j])%2==0){
                flag=0;
                break;
            }

            if(j>=i){
                sum-=f[j-i+1];
            }
        }

        if(flag){
            if(res>ans){
                res=ans;
                resx=i;
            }
        }

//        for(int j=1;j<=n;j++){
//            cout<<f[j]<<" ";
//        }
//        cout<<endl;




    }



    printf("%d %d",resx,res);


    return 0;
}