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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4052 [JSOI2007]文本生成器

P4052 [JSOI2007]文本生成器

AC自动机套路题了。

26个字母,任意生成一个长度为 $n$ 的串,求有多少个串包含至少一个给定的 $n$ 个串。

考虑计算多少个串不包含任意一个串,建立后缀自动机,做DP,当且仅当不能构成一个串时转移。

查看代码
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#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int N = 6e3 + 10, M = 110, mod = 1e4 + 7;
char s[N];
int n, m, f[N][M];
int idx, cnt[N], nxt[N], tr[N][26];
int binpow (int b, int k)
{
int res = 1;
while (k)
{
if (k & 1)
res = res * b % mod;
b = b * b % mod;
k >>= 1;
}
return res;
}
void insert (int len, char *str)
{
int p = 0;
for (int i = 1; i <= len; i++)
{
int &t = tr[p][str[i] - 'A'];
!t && (t = ++idx);
p = t;
}
cnt[p]++;
}
void build ()
{
queue <int> q;
for (int i = 0; i < 26; i++)
tr[0][i] && (q.push(tr[0][i]), 0);
while (!q.empty())
{
int t = q.front();
q.pop();
for (int i = 0; i < 26; i++)
{
int &p = tr[t][i];
if (!p)
p = tr[nxt[t]][i];
else
{
nxt[p] = tr[nxt[t]][i];
q.push(p);
}
}
cnt[t] += cnt[nxt[t]];
}
}
int main ()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
{
scanf("%s", s + 1);
insert(strlen(s + 1), s);
}
build();
f[0][0] = 1;
for (int k = 0; k < m; k++)
for (int i = 0; i <= idx; i++)
for (int j = 0; j < 26; j++)
if (!cnt[tr[i][j]])
(f[tr[i][j]][k + 1] += f[i][k]) %= mod;
int res = binpow(26, m);
for (int i = 0; i <= idx; i++)
(res -= f[i][m]) %= mod;
printf("%d", (res + mod) % mod);
return 0;
}