poj3276 Face The Right Way 反转问题

Face The Right Way

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5709 Accepted: 2641

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

题意：给n头牛的站位：B-向后，F向前。每次反转连续的k个，使其全部向前，求反转次数最小值m及其对应的k值。

分析：

设f[i]为[i,i+k-1]这个区间进行反转，对于第i头牛，只需统计sum = f[i-k+1]--f[i-1]反转的次数，结合本身牛的占位，就可以决定第i头牛是否需要来一次反转。

而对于下一头牛i+1，需要统计sum1 = f[i+1-k+1]--f[i] = sum+f[i]-f[i-k+1]。设置sum变量维护这个数值，可以将每次将f[i]用常熟时间计算出来。

#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INF 0x3f3f3f3f#define MAX 100typedef long long ll;int n;int dir[5001];int f[5001];int K,M;int cal(int k){//计算前n-k+1个值的f[i]memset(f,0,sizeof(f));int sum = 0;int res = 0;for (int i = 0; i+k-1 < n; ++i){if((sum+dir[i])%2 != 0){res++;f[i] = 1;} sum += f[i];if(i-k+1 >= 0) sum -= f[i-k+1];}//检验后面剩余的值for (int i = n-k+1; i < n; ++i){if((sum+dir[i])%2 != 0) return -1;if(i-k+1 >= 0) sum -= f[i-k+1]; } return res;}void solve(){M = n;for (int k = 1; k <= n; ++k){int m = cal(k);if(m>=0 && M>=m) K = k,M = m;}printf("%d %d\n",K,M);}int main(int argc, char const *argv[]){scanf("%d",&n);for (int i = 0; i < n; ++i){char c;cin >> c;if(c == 'B') dir[i] = 1;//向后的为1else dir[i] = 0;//向前为0}solve();}