POJ 3276 Face The Right Way
阿新 • • 發佈:2018-10-15
contain ini ber set else if cow name NPU war (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.
, the minimum number of machine operations required to get all the cows facing forward using that value of K.
Lines 2..N+1: Line i+1 contains a single character, F or B, indicating whether cow i is facing forward or backward.
http://poj.org/problem?id=3276
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
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
Input
Line 1: A single integer: NLines 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 MSample Input
7 B B F B F B B
Sample Output
3 3
時間復雜度:$O(N ^ 2)$
代碼:
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; const int maxn = 1e5 + 10; int N; int dir[maxn], f[maxn]; int cal(int K) { memset(f, 0, sizeof(f)); int res = 0, sum = 0; for(int i = 0; i + K <= N; i ++) { if((dir[i] + sum) % 2 != 0) { 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) % 2 != 0) return -1; if(i - K + 1 >= 0) sum -= f[i - K + 1]; } return res; } void solve() { int K = 1, M = N; for(int k = 1; k <= N; k ++) { int m = cal(k); if(m >= 0 && M > m) { M = m; K = k; } } printf("%d %d\n", K, M); } int main() { scanf("%d", &N); for(int i = 0; i < N; i ++) { char a[2]; scanf("%s", a); if(strcmp(a, "F") == 0) dir[i] = 0; else if(strcmp(a, "B") == 0) dir[i] = 1; } //for(int i = 0; i < N; i ++) //printf("%d\n", dir[i]); solve(); return 0; }
POJ 3276 Face The Right Way