Description
Farmer John has arranged hisN(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 turnK(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 ofKcows in the line (one cannot use it on fewer thanKcows, 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 ofK, please help him determine the minimum value ofKthat minimizes the number of operations required by the machine to make all the cows face forward. Also determineM, the minimum number of machine operations required to get all the cows facing forward using that value ofK.
Input
Line 1: A single integer:NLines 2..N+1: Linei+1 contains a single character,ForB, indicating whether cowiis facing forward or backward.
Output
Line 1: Two space-separated integers:KandM
Sample Input
7BBFBFBB
Sample Output
3 3
Hint
ForK= 3, the machine must be operated three times: turn cows (1,2,3), (3,4,5), and finally (5,6,7)
首先枚举k，对于一个特定的k，最左端只有一个区间能够影响得到，所以可以确定是否反转。然后这个端点就不会再被影响，这样长度就减1了，重复以上步骤就能得出答案。还运用到了一个小技巧，用f[i]来记录i是否被反转，这样就不需要真正的实现反转，时间复杂度降低。
#include <iostream>#include <sstream>#include <fstream>#include <string>#include <map>#include <vector>#include <list>#include <set>#include <stack>#include <queue>#include <deque>#include <algorithm>#include <functional>#include <iomanip>#include <limits>#include <new>#include <utility>#include <iterator>#include <cstdio>#include <cstdlib>#include <cstring>#include <cctype>#include <cmath>#include <ctime>using namespace std;int n;int dir[5010];//牛的方向(0:F, 1:B)int f[5010];//区间[i, i+k-1]是否被反转int calc(int k){memset(f, 0, sizeof(f));int res = 0, sum = 0;for (int i = 0; i + k <= n; ++i){if ((dir[i]+sum)&1){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)&1)return -1;if (i – k + 1 >= 0)sum -= f[i-k+1];}return res;}int main(){cin >> n;char s[5];for (int i = 0; i < n; ++i){scanf("%s", s);dir[i] = (s[0] == 'B');}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;}}cout << K << ' ' << M << endl;return 0;}
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你并不一定会从此拥有更美好的人生，
