AtCoder Beginner Contest 286——C - Rotate and Palindrome

AtCoder Beginner Contest 286 题目讲解

A题 B题 C题 D题 E题

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蒟蒻来讲题，还望大家喜。若哪有问题，大家尽可提！

Hello, 大家好哇！本初中生蒟蒻今天讲解一下AtCoder Beginner Contest 286的C题——Rotate and Palindrome

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原题

Problem Statement

You are given a string $S$ of length $N$. Let $S_i(1\le i\le N)$ be the $i$-th character of $S$ from the left.

You may perform the following two kinds of operations zero or more times in any order:

• Pay $A$ yen (the currency in Japan). Move the leftmost character of S to the right end. In other words, change $S_1{S}_2\dots S_N$​ to $S_2\dots S_N{S}_1$.
• Pay $B$ yen. Choose an integer $i$ between $1$ and $N$, and replace $S_i$​ with any lowercase English letter.

How many yen do you need to pay to make $S$ a palindrome?
Note:
Palindrome is a string $T$ is a palindrome if and only if the $i$-th character from the left and the $i$-th character from the right are the same for all integers $i(1\le i\le \left|T\right|)$, where $\left|T\right|$ is the length of $T$.

Constraints

• $1\le N\le 5000$
• $1\le A,B\le 10^9$
• $S$ is a string of length $N$ consisting of lowercase English letters.
• All values in the input except for $S$ are integers.

Input
The input is given from Standard Input in the following format:

N  A  B
S

Output

Print the answer as an integer.

Sample Input 1

5 1 2
rrefa

Sample Output 1

3

First, pay $2$ yen to perform the operation of the second kind once: let $i=5$ to replace $S_5$​ with $e$. S is now $rrefe$.
Then, pay $1$ yen to perform the operation of the first kind once.
$S$ is now refer, which is a palindrome.
Thus, you can make $S$ a palindrome for $3$ yen. Since you cannot make $S$ a palindrome for $2$ yen or less, $3$ is the answer.

Sample Input 2

8 1000000000 1000000000
bcdfcgaa

Sample Output 2

4000000000

Note that the answer may not fit into a $32$-bit integer type.

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思路

本题也其实是一个暴力枚举，我们枚举每一次进行n次1操作（将S的第一个放入最后一个位置），对于每一次我们判断最少的花费即可。
最小的花费算法：
我们因为对于每一个操作了$1,2,...,n$的$S$，那么$S$变成操作最少的回文串是$S_1,...,S_{\lfloor \frac{n}{2}\rfloor +(n\%2\ne 0)},S_{\lfloor \frac{n}{2}\rfloor },S_{\lfloor \frac{n}{2}\rfloor -1},...,S_1$

就比如说：
对于字符串$abcda$，它的最少操作的回文串是$abcba$

因此，构造出回文串，在对于每一个字符进行比较计算即可！

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代码

#include 
#define int long long //本题需要开long long！

using namespace std;

int n, a, b;
deque<char> past, match;

int cost()
{
	//构造匹配的回文串
	match.clear();
	for (int i = 1; i <= n / 2 + (n & 1); i ++)
		match.push_back(past[i]);
	for (int i = n / 2; i >= 1; i --)
		match.push_back(past[i]);
	
	match.push_front('0');
	//判断每一个字符是否相同，计算花费
	int num = 0;
	for (int i = 1; i <= n; i ++)
		num += (bool(match[i] != past[i]) * b);
	
	return num;
}

signed main()
{
	cin.tie(0);
    ios::sync_with_stdio(false);
	
	cin >> n >> a >> b;
	
	char input;
	
	past.push_back('0');
	for (int i = 1; i <= n; i ++)
		cin >> input, past.push_back(input);
	
	int ans = 1e18;
	
	ans = min(ans, cost());
	for (int i = 1; i <= n; i ++)
	{
		past.push_back(past[1]), past.pop_front();
		past.pop_front();
		past.push_front('0');
		ans = min(ans, cost() + i * a);//不停枚举操作1，找最小值
	}
	
	cout << ans << endl;
	return 0;
}

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今天就到这里了！

大家有什么问题尽管提，我都会尽力回答的！最后，除夕夜祝大家新年快乐！

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