AtCoder备赛刷题 ABC 363 | Avoid Palindrome 2

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附上汇总贴：AtCoder 备赛刷题 | 汇总

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【Problem Statement】

You are given a string $S$ of length $N$ consisting only of lowercase English letters.

给定一个长度为 $N$ 的字符串 $S$，仅由小写英文字母组成。

Find the number of strings obtained by permuting the characters of $S$ (including the string $S$ itself) that do not contain a palindrome of length $K$ as a substring.

查找通过排列 $S$ 的字符（包括字符串 $S$ 本身）而获得的字符串数量，其中不包含长度为 $K$ 的回文作为子字符串。

Here, a string $T$ of length $N$ is said to “contain a palindrome of length $K$ as a substring” if and only if there exists a non-negative integer ii not greater than $(N-K)$ such that $T_{i+j}=T_{i+K+1-j}$ for every integer $j$ with $1\le j\le K$.

这里，长度为 $N$ 的字符串 $T$ 被称为“包含长度为 $K$ 的回文作为子字符串”，当且仅当存在一个不大于 $(N-K)$ 的非负整数 $i$，使得对于每个整数 $j$，$T_{i+j}=T_{i+K+1-j}$ ，且$1\le j\le K$。

Here, $T_k$ denotes the $k$-th character of the string $T$.

这里，$T_k$ 表示字符串 $T$ 的第 $k$ 个字符。

【Constraints】

• $2\le K\le N\le 10$
• $N$ and $K$ are integers.
• $S$ is a string of length $N$ consisting only of lowercase English letters.

【Input】

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

N K 
S

【Output】

Print the number of strings obtained by permuting $S$ that do not contain a palindrome of length $K$ as a substring.

【Sample Input 1】

3 2
aab

【Sample Output 1】

1

The strings obtained by permuting aab are aab, aba, and baa. Among these, aab and baa contain the palindrome aa of length $2$ as a substring.
Thus, the only string that satisfies the condition is aba, so print $1$.

【Sample Input 2】

5 3
zzyyx

【Sample Output 2】

16

There are $30$ strings obtained by permuting zzyyx, $16$ of which do not contain a palindrome of length $3$. Thus, print $16$.

【Sample Input 3】

10 5
abcwxyzyxw

【Sample Output 3】

440640

【代码详解】

《AtCoder Avoid K Palindrome 2》 #枚举#

#include 
using namespace std;
const int N = 15;
int n, k, ans;
bool st[N];
char a[N], path[N];
map<string, int> mp;
int check()
{
    for (int i=0; i+k-1<n; i++)
    {
        bool flag = true;
        int j = i+k-1;
        for (int x=i, y=j; x<y; x++, y--)
            if (path[x]!=path[y]) flag = false;
        if (flag) return 0;
    }
    return 1;
}
void dfs(int u)
{
    if (u==n)
    {
        string s="";
        for (int i=0; i<n; i++)
            s += path[i];
        if (!mp[s]) 
        {
            mp[s] = 1;
            ans += check();
        }
        return;
    }
    for (int i=0; i<n; i++)
    {
        if (!st[i])
        {
            path[u] = a[i];
            st[i] = true;
            dfs(u+1);
            st[i] = false;
        }
    }
}
int main()
{
    cin >> n >> k;
    for (int i=0; i<n; i++)
        cin >> a[i];
    sort(a, a+n);
    dfs(0);
    cout << ans << endl;
    return 0;
}

【运行结果】

3 2
aab
1