P2882 [USACO07MAR]Face The Right Way G 【贪心 + 差分】

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，求操作的最少 次数M和对应的最小K。
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

7
B
B
F
B
F
B
B

输出 #1

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)

枚举K，用差分修改一段区间，时间复杂度为n^2。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 1e4 + 10;
int a[maxn];
int vis[maxn];
int n;

int main() {
	scanf("%d", &n);
	getchar();
	char c;
	for(int i = 1; i <= n; i++) {
		scanf("%c", &c);
		getchar();
		a[i] = c == 'F';
	}
	int step = 1e9, len = 0;
	for(int k = 1 ; k <= n; k++) {
		memset(vis, 0, sizeof(vis));
		int tem = 0, f = 0, tag = 0;
		for(int i = 1; i <= n; i++) {
			if(vis[i])	tag--;
			if((a[i] + tag) % 2 == 0) {
				if(i + k - 1 <= n) {
					tag++, vis[i + k] = true;
					tem++;
				}
				else	f = true;
			}
		//	cout << tag << " ";
		}
//		cout << "   " << k << endl;
		if(f == 0 && tem < step)	step = tem, len = k;
	}
	printf("%d %d", len, step);
	return 0;
}