洛谷P2882 [USACO07MAR]Face The Right Way G 贪心思想的数学证明

牛的方向翻转优化

最新推荐文章于 2024-10-25 14:58:48 发布

原创最新推荐文章于 2024-10-25 14:58:48 发布 · 384 阅读

1 ·

CC 4.0 BY-SA版权

OI 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了如何通过贪心策略解决一个关于翻转牛的方向的问题，以使所有牛都面向前方，同时给出了数学证明和代码实现。算法核心是每次翻转以遇到的第一个向后牛为首的K头牛，通过枚举K的值找到最小操作次数。

题目

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 KK 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
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)

题解

（引自Delva）

贪心，枚举，差分

因为同一个点翻转两次就与没有翻转的效果相同了，因此我们有一个贪心策略为：

从左到右对于出现的每一个B翻转一次从当前点开始的区间，就能保证是最优解。

样例的模拟：

如果当前翻转的区间长度为3

粗体表示当前下标

B B F B F B B

(B B F) B F B B

F F B B F B B

F F B B F B B

F F B B F B B

F F (B B F) B B

F F F F B B B

F F F F B B B

F F F F B B B

F F F F (B B B)

F F F F F F F

F F F F F F F

F F F F F F F

但是我们会发现这样是 $n^2$ 的，再枚举长度，就变为了 $n^3$ .因此，需要对区间翻转差分一下，总时间复杂度就是 $n^2$ 的了。

代码（自己实现）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

#ifndef INT_MAX
#define INT_MAX 0x7fffffff
#endif

using namespace std;

const int MAXN = 5005;
int N, M = INT_MAX, ans;
bool a[MAXN], b[MAXN];

int main()
{
    cin >> N;
    for(int i = 1; i <= N; ++i)
    {
        char ch;
        cin >> ch;
        a[i] = ch == 'F';
    }
    for(int K = 1; K <= N; ++K)
    {
        memset(b, 0, sizeof(b)); // b：差分数组
        bool succ = true, g = false;
        int cnt = 0;
        for(int i = 1; i <= N; ++i)
        {
            g ^= b[i]; // '^'相当于'+'或'-'
            if((a[i] ^ g) == false)
            {
                if(i + K - 1 > N)
                {
                    succ = false;
                    break;
                }
                g ^= 1;
                b[i + K] ^= 1;
                ++cnt;
            }
        }
        if(succ)
        {
            if(cnt < M)
            {
                M = cnt;
                ans = K;
            }
        }
    }
    printf("%d %d\n", ans, M);
    return 0;
}