題目鏈接

思路

假設讓求長度為$L$且不包含詞根的個數，對所有的詞根建$ac$自動機，然后用矩陣$M$表示可轉移狀態，最后最快速冪即可。
如何求長度不超過$L$且不包含$L$且不包含詞根的個數，可以利用這個矩陣$\begin{array}{c}\left[\begin{array}{cc}M& 1\\ 0& 1\end{array}\right]\end{array}$，這樣就可以得到$\sum _{i=0}^n{M}^i$，最后用總的狀態數${\left[\begin{array}{cc}26& 1\\ 0& 1\end{array}\right]}^n$減去即可

#include <vector> #include <queue> #include <stdio.h> #include <iostream> const int maxn = 1e5 + 5; const int inf = 0x3f3f3f3f; using namespace std; struct Matrix{int n;unsigned long long mat[151][151];Matrix(){}Matrix(int _n) {n = _n;for (int i = 0; i < n; ++i) {for (int j = 0; j < n; ++j) {mat[i][j] = 0;}}}Matrix operator *(const Matrix &b) const{Matrix ans = Matrix(n);for (int i = 0; i < n; ++i) {for (int j = 0; j < n; ++j) {for (int k = 0; k < n; ++k) {ans.mat[i][j] += mat[i][k] * b.mat[k][j];// ans.mat[i][j] = (ans.mat[i][j] + mat[i][k] * b.mat[k][j] % mod) % mod;}}}return ans;} }; struct Trie{int nex[maxn][26], fail[maxn], end[maxn];int root, p;inline int newnode() {for (int i = 0; i < 26; ++i) {nex[p][i] = -1;}end[p++] = 0;return p - 1;}inline void init() {p = 0;root = newnode();}inline void insert(char *buf) {int now = root;for (int i = 0; buf[i]; ++i) {if (nex[now][buf[i]-'a'] == -1) nex[now][buf[i]-'a'] = newnode();now = nex[now][buf[i]-'a'];}end[now]++;} inline void build() {queue<int> que;fail[root] = root;for (int i = 0; i < 26; ++i) {if (nex[root][i] == -1)nex[root][i] = root;else {fail[nex[root][i]] = root;que.push(nex[root][i]);}}while (!que.empty()) {int now = que.front();que.pop();if (end[fail[now]]) end[now] = 1; for (int i = 0; i < 26; ++i) {if (nex[now][i] == -1) nex[now][i] = nex[fail[now]][i];else {fail[nex[now][i]] = nex[fail[now]][i];que.push(nex[now][i]);}}}}inline Matrix get() {Matrix tmp = Matrix(p*2);for (int i = 0; i < p; ++i) {for (int j = 0; j < 26; ++j) {if (end[i] || end[nex[i][j]]) continue;tmp.mat[i][nex[i][j]]++;}}for (int i = 0; i < p; ++i) tmp.mat[i][i+p] = tmp.mat[i+p][i+p] = 1;return tmp;} }ac; Matrix Pow(Matrix a, int n) {Matrix ans = Matrix(a.n);Matrix tmp = a;for (int i = 0; i < a.n; ++i) {ans.mat[i][i] = 1;}while (n) {if (n & 1) ans = ans * tmp;tmp = tmp * tmp;n >>= 1;}return ans; } char s[maxn]; int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int n, m;while (scanf("%d %d\n", &n, &m) != EOF) {ac.init();for (int i = 0; i < n; ++i) scanf("%s", s), ac.insert(s);ac.build();Matrix ans = ac.get();ans = Pow(ans, m);Matrix tmp = Matrix(2);tmp.mat[0][0] = 26;tmp.mat[0][1] = 1;tmp.mat[1][1] = 1;tmp = Pow(tmp, m);unsigned long long sum = tmp.mat[0][0] + tmp.mat[0][1] - 1;for (int j = 0; j < 2*ac.p; ++j) sum -= ans.mat[0][j];printf("%llu\n", sum+1);}return 0; }

總結

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