《P2882 [USACO07MAR] Face The Right Way G》

题目描述

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.

N 头牛排成一列 1≤N≤5000。每头牛或者向前或者向后。为了让所有牛都面向前方，农夫每次可以将 K 头连续的牛转向 1≤K≤N，求使操作次数最小的相应 K 和最小的操作次数 M。F 为朝前，B 为朝后。

请在一行输出两个数字 K 和 M，用空格分开。

输入格式

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.

输出格式

Line 1: Two space-separated integers: K and M

输入输出样例

输入 #1复制

7
B
B
F
B
F
B
B

输出 #1复制

3 3

说明/提示

For K = 3, the machine must be operated three times: turn cows (1,2,3), (3,4,5), and finally (5,6,7)

代码实现：

#include <cstdio>
#include <cstring>

using namespace std;

// 表示元素个数，将n改为更具表意性的elementCount
int elementCount, 
    // 用于存储原始数据，将a改为originalData
    originalData[5005], 
    // 记录最小操作步数，将minn改为minSteps，初始值根据实际意义调整为合适较大值
    minSteps = 1e9, 
    // 记录当前操作步数，将step改为currentSteps
    currentSteps, 
    // 记录起始位置，将pin改为startPosition
    startPosition, 
    // 用于存储差异数据，将dif改为differenceData
    differenceData[5005], 
    // 记录当前状态值，将now改为currentValue
    currentValue, 
    // 临时存储数组，将b改为tempArray
    tempArray[5005];

int main() {
    scanf("%d", &elementCount);
    for (int i = 1; i <= elementCount; i++) {
        char ch;
        getchar();
        ch = getchar();
        if (ch == 'B') 
            originalData[i] = 0;
        else 
            originalData[i] = 1;
    }
    for (int i = 1; i <= elementCount; i++) {
        memset(tempArray, 0, sizeof(tempArray));
        currentValue = 0;
        currentSteps = 0;
        int flag = 0;
        for (int j = 1; j <= elementCount; j++)
            differenceData[j] = originalData[j];
        for (int j = 1; j <= elementCount - i + 1; j++) {
            currentValue ^= tempArray[j];
            differenceData[j] ^= currentValue;
            if (!differenceData[j]) {
                currentValue ^= 1;
                tempArray[i + j] ^= 1;
                currentSteps++;
            }
        }
        for (int j = elementCount - i + 2; j <= elementCount; j++) {
            currentValue ^= tempArray[j];
            if (!(differenceData[j] ^ currentValue)) {
                flag = 1;
                break;
            }
        }
        if (!flag) {
            if (minSteps > currentSteps) {
                minSteps = currentSteps;
                startPosition = i;
            }
        }
    }
    printf("%d %d", startPosition, minSteps);
    return 0;
}