hdu 2243(ac自动机+矩阵快速幂)

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题意：有n个小写字母组成的模式串，问长度不超过L的小写字母串中至少出现一个模式串的种类是多少。
题解：这道题和poj 2778类似，不过是把长度小于L的串的可能情况也计入，把邻接矩阵多一维存总和，然后结果用总种类数减一个模式串也不出现的种类数。总种类数很大，26^1 + 26^2 + 26^3 + … + 26^n，也用矩阵快速幂计算。f(n) = 26 * f(n - 1) + 26。
初始矩阵：
| f(n - 1) 1 |
| 0 0 |
系数矩阵：
| 26 0 |
| 26 1 |
相乘得到：
| f(n) 1 |
| 0 0 |

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>#define ll unsigned long longusing namespace std;const int N = 55;const int SIGMA_SIZE = 26;struct Mat {    ll a[N][N];}ori, res;int Next[N][SIGMA_SIZE], fail[N], val[N], sz, n, L;char str[N];void init() {    sz = 1;    memset(Next[0], 0, sizeof(Next[0]));    val[0] = 0;}void insert(char *s) {    int u = 0, len = strlen(s);    for (int i = 0; i < len; i++) {        int k = s[i] - 'a';        if (!Next[u][k]) {            memset(Next[sz], 0, sizeof(Next[sz]));            val[sz] = 0;            Next[u][k] = sz++;        }        u = Next[u][k];    }    val[u] = 1;}void getFail() {    queue<int> Q;    fail[0] = 0;    for (int i = 0; i < SIGMA_SIZE; i++)        if (Next[0][i]) {            fail[Next[0][i]] = 0;            Q.push(Next[0][i]);        }    while (!Q.empty()) {        int u = Q.front();        Q.pop();        if (val[fail[u]])            val[u] = 1;        for (int i = 0; i < SIGMA_SIZE; i++) {            if (!Next[u][i])                Next[u][i] = Next[fail[u]][i];            else {                fail[Next[u][i]] = Next[fail[u]][i];                Q.push(Next[u][i]);            }        }    }}Mat multiply(const Mat &x, const Mat &y) {    Mat temp;    for (int i = 0; i <= sz; i++)        for (int j = 0; j <= sz; j++) {            temp.a[i][j] = 0;            for (int k = 0; k <= sz; k++)                temp.a[i][j] += x.a[i][k] * y.a[k][j];        }    return temp;}void calc(int m) {    while (m) {        if (m & 1)            res = multiply(res, ori);        m >>= 1;        ori = multiply(ori, ori);    }}int main() {    while (scanf("%d%d", &n, &L) == 2) {        init();        for (int i = 0; i < n; i++) {            scanf("%s", str);            insert(str);        }        getFail();        for (int i = 0; i <= sz; i++)            for (int j = 0; j <= sz; j++)                res.a[i][j] = ori.a[i][j] = 0;        for (int i = 0; i <= sz; i++)            res.a[i][i] = 1;        for (int i = 0; i < sz; i++)            for (int j = 0; j < SIGMA_SIZE; j++)                if (!val[Next[i][j]])                    ori.a[i][Next[i][j]]++;        for (int i = 0; i <= sz; i++)            ori.a[i][sz] = 1;        calc(L);        ll ans = 0;        for (int i = 0; i <= sz; i++)            ans += res.a[0][i];        ori.a[0][0] = ori.a[1][0] = 26;        ori.a[0][1] = 0;        ori.a[1][1] = 1;        res.a[0][1] = 1;        res.a[0][0] = res.a[1][1] = res.a[1][0] = 0;        sz = 1;        calc(L);        ll ans2 = res.a[0][0];        printf("%llu\n", ans2 - ans + 1);    }    return 0;}

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