考研路茫茫――单词情结
AC自动机+矩阵快速幂好题
题目描述
发现 L 很大，考虑矩阵快速幂，先考虑朴素，求出答案的补集，设 \(f_{i,j}\)，\(i\) 表示当前在单词的哪一位上，\(j\) 表示在 Fail 树上的哪一位上，不接触有单词的地方（非得用答案补集的原因是直接考虑答案太麻烦，得要高深的容斥），可得转移式 if (!t[t[j].son[c]].end) f[i + 1][t[j].son[c]] += f[i][j];。
最终答案即为：\(Ans=\sum26^k-\sum\limits_{0\le i\le L,0\le j\le cnt} f_{i,j}\)
前者可以用动态规划求出 \(g_{i}=26g_{i-1}+1\)
那么可以有这个代码：
#include<bits/stdc++.h>
#define LL long long
//#define int LL
#define per(i, a, b) for (int i = a, END##i = b; i >= END##i; i--)
#define rep(i, a, b) for (int i = a, END##i = b; i <= END##i; i++)
#define repn(x) rep(x, 1, n)
#define repm(x) rep(x, 1, m)
#define pb push_back
#define PII pair<int, int>
#define i64 unsigned long long
#define YY puts("Yes"), exit(0)
#define NN puts("No"), exit(0)
using namespace std;
const int Mod = 1e9 + 7;
const int Inf = 0x3f3f3f3f;
const LL InfLL = 0x3f3f3f3f3f3f3f3f;
inline LL read() {LL s = 0, fu = 1; char ch = getchar(); while (ch < '0' || ch > '9') ch == '-' ? fu = -1 : 0, ch = getchar(); while (ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * fu;}
const int N = 110;
int n, L;
struct Node {
int son[26], fail;
bool end;
}t[N]; int cnt;
char str[N];
void Insert(char *str) {
int now = 0;
for (int i = 0; str[i]; i++) {
int w = str[i] - 'a';
if (!t[now].son[w]) t[now].son[w] = ++cnt;
now = t[now].son[w];
}
t[now].end = 1;
}
void getFail() {
queue<int> q;
rep(i, 0, 25) if (t[0].son[i])
q.push(t[0].son[i]);
while (!q.empty()) {
int now = q.front();
q.pop(); t[now].end |= t[t[now].fail].end;
rep(i, 0, 25) {
if (t[now].son[i]) {
t[t[now].son[i]].fail = t[t[now].fail].son[i];
q.push(t[now].son[i]);
} else t[now].son[i] = t[t[now].fail].son[i];
}
}
}
i64 f[N][N], g[N];
i64 qpow(i64 a, int b) {
i64 ans = 1;
while (b) {
if (b & 1) ans = a * ans;
a = a * a;
b >>= 1;
}
return ans;
}
inline void Main() {
while (scanf("%d%d", &n, &L) != -1) {
cnt = 0; memset(t, 0, sizeof(t));
while (n--) {
scanf("%s", str);
Insert(str);
}
getFail();
memset(f, 0, sizeof(f));
f[0][0] = 1;
rep(i, 0, L - 1) rep(j, 0, cnt) rep(c, 0, 25)
if (!t[t[j].son[c]].end) f[i + 1][t[j].son[c]] += f[i][j];
g[0] = 1; i64 ans = 0;
rep(i, 1, L) g[i] = 26 * g[i - 1] + 1;
rep(i, 0, L) rep(j, 0, cnt)
ans += f[i][j];
// cout << g[1] << " " << g[2] << " " << g[3] << "\n";
cout << g[L] - ans << "\n";
}
}
signed main() {
// freopen("input.in", "r", stdin);
int T = 1;
while (T--)
Main();
return 0;
}
考虑优化，首先优化简单的部分，即 \(\sum26^k\) 可得矩阵：
\[\begin{bmatrix}
G(n) \\
1
\end{bmatrix}
=
\begin{bmatrix}
26 & 1 \\
0 & 1
\end{bmatrix}
\times
\begin{bmatrix}
G(n-1) \\
1
\end{bmatrix}
\]
观察转移式 if (!t[t[j].son[c]].end) f[i + 1][t[j].son[c]] += f[i][j]; ，可以发现可以转化成在 Fail 树上连边（可以连的），然后求值，也就是可以将这个转移过程改成，从 Fail 树的根节点往下转移，重复 L 次，且符合条件，t[j].son[c] 到 j 的路径有多少个可以转化成 \(g_{i,j}=\sum g_{i,k}g_{k,j}\) 可以发现是矩阵快速幂的公式。
\(\mathscr{Code:}\)
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define LL long long
//#define int LL
#define per(i, a, b) for (int i = a, END##i = b; i >= END##i; i--)
#define rep(i, a, b) for (int i = a, END##i = b; i <= END##i; i++)
#define repn(x) rep(x, 1, n)
#define repm(x) rep(x, 1, m)
#define pb push_back
#define PII pair<int, int>
#define i64 unsigned long long
#define YY puts("Yes"), exit(0)
#define NN puts("No"), exit(0)
using namespace std;
const int Mod = 1e9 + 7;
const int Inf = 0x3f3f3f3f;
const LL InfLL = 0x3f3f3f3f3f3f3f3f;
inline LL read() {LL s = 0, fu = 1; char ch = getchar(); while (ch < '0' || ch > '9') ch == '-' ? fu = -1 : 0, ch = getchar(); while (ch >= '0' && ch <= '9') s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar(); return s * fu;}
const int N = 50;
int n;
i64 L;
struct Node {
int son[26], fail;
bool end;
}t[N]; int cnt;
char str[N];
void Insert(char *str) {
int now = 0;
for (int i = 0; str[i]; i++) {
int w = str[i] - 'a';
if (!t[now].son[w]) t[now].son[w] = cnt++;
now = t[now].son[w];
}
t[now].end = 1;
}
void getFail() {
queue<int> q;
rep(i, 0, 25) if (t[0].son[i])
q.push(t[0].son[i]);
while (!q.empty()) {
int now = q.front();
q.pop(); t[now].end |= t[t[now].fail].end;
rep(i, 0, 25) {
if (t[now].son[i]) {
t[t[now].son[i]].fail = t[t[now].fail].son[i];
q.push(t[now].son[i]);
} else t[now].son[i] = t[t[now].fail].son[i];
}
}
}
struct Matrix {
i64 M[N][N], n;
Matrix() {};
Matrix(int _) {
n = _;
memset(M, 0, sizeof(M));
}
Matrix operator* (const Matrix& b) const {
Matrix res = Matrix(n);
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
for (int k = 0; k < n; k++)
res.M[i][j] += M[i][k] * b.M[k][j];
return res;
}
};
Matrix qpow(Matrix a, i64 b) {
Matrix res = Matrix(a.n);
rep(i, 0, a.n - 1) res.M[i][i] = 1;
repn(i) res.M[i][i] = 1;
while (b) {
if (b & 1) res = a * res;
a = a * a;
b >>= 1;
}
return res;
}
inline void Main() {
while (cin >> n >> L) {
cnt = 1; memset(t, 0, sizeof(t));
while (n--) {
scanf("%s", str);
Insert(str);
}
getFail();
Matrix M1 = Matrix(2);
M1.M[0][0] = 26, M1.M[1][1] = M1.M[0][1] = 1;
M1 = qpow(M1, L);
i64 ans = M1.M[0][0] + M1.M[0][1];
Matrix M2 = Matrix(cnt + 1);
rep(i, 0, cnt - 1) rep(j, 0, 25) {
if (t[t[i].son[j]].end) continue;
M2.M[i][t[i].son[j]]++;
}
for (int i = 0; i < cnt + 1; i++)
M2.M[i][cnt] = 1;
M2 = qpow(M2, L);
rep(i, 0, M2.n - 1) ans -= M2.M[0][i];
cout << ans << "\n";
}
}
signed main() {
// freopen("input.in", "r", stdin);
int T = 1;
while (T--)
Main();
return 0;
}
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