poj 2778 AC自动机构建有向图 + 邻接矩阵快速幂

Problem:
给你m个病毒串,求指定长度n且不含病毒串作为子串的字符串一共有多少种.
Analyse:
用AC自动机构建L个状态节点,每个节点的end标记记录是否在这里形成病毒串.
这里有个核心就是,如果当前后缀的子后缀(也就是它的fail指针指向的地方)是病毒串的话,
那么它就是病毒串.
然后根据这个AC自动机的L个节点来建立有向图的邻接矩阵B,B[i][j]代表从i到j状态的路径数量.
B[0][j]代表的是从初始状态,还没有任何字符的时候转移到j状态,因为根节点0就是没有任何限制.
然后Bk之后,B[i][j]代表的是,从i到j长度为k的路径有多少条.
我们需要求得就是从根节点出发到任意节点,长度为n的路径的条数.

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 一  5/ 9 11:49:40 2016 *File Name*     : poj2778.cpp**Code**:***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a));#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e5) + 0, INF = 0x3fffffff;const int maxn = 1e5 + 13;const int maxtrie = 11 * 10 + 12; //注意这里为最多500个单词,每个单词最多500个字母,所以节点最多为乘const int maxcharset = 4; //字符集合const int charst = 0;char buf[11];class matrix{public:    ll M[maxtrie][maxtrie], n;    matrix (int N){        n = N;        for (int i = 0; i < N; i++)             for (int j = 0; j < N; j++)                M[i][j] = 0;    }    matrix(int N, int x) {        n = N;        for (int i = 0; i < N; i++) M[i][i] = x;    }    matrix operator* (const matrix &b) {        matrix &a = *this;        matrix ret(b.n);        for (int i = 0; i < n; i++) {            for (int j = 0; j < n; j++) {                ret.M[i][j] = 0;                for (int k = 0; k < n; k++) {                    ret.M[i][j] += a.M[i][k] * b.M[k][j]  % mod;                }                ret.M[i][j] = ret.M[i][j] % mod;            }        }        return ret;    }    matrix operator^ (int b) {        matrix ret(this -> n, 1);        matrix a = *this;        while (b) {            if (b & 1) ret = ret * a;            a = a * a;            b >>= 1;        }        return ret;    }};struct Trie{    int next[maxtrie][maxcharset], fail[maxtrie], end[maxtrie];    int root, L;    int newnode(void) {        for (int i = 0;i < maxcharset;i++) next[L][i] = 0;        end[L++] = 0;        return L - 1;    }    int getid(char c) {        if (c == 'A') return 0;        if (c == 'T') return 1;        if (c == 'C') return 2;        if (c == 'G') return 3;        return 0;    }    void init(void) {        L = 0;        root = newnode();    }    void insert(char buf[]) {        int len = (int)strlen(buf);        int now = root;        for(int i = 0; i < len; i++) {            if(next[now][getid(buf[i])] == 0)                next[now][getid(buf[i])] = newnode();            now = next[now][getid(buf[i])];        }        end[now] = true;    }    void build(void) {        queue<int> Q;        fail[root] = root;        for(int i = 0;i < maxcharset;i++)            if(next[root][i] == 0) next[root][i] = root;            else {                fail[next[root][i]] = root;                Q.push(next[root][i]);            }        while (!Q.empty())  {            int now = Q.front(); Q.pop();            for(int i = 0;i < maxcharset;i++)                if(next[now][i] == 0) {                    next[now][i] = next[fail[now]][i];                    end[now] = end[now] || end[fail[now]];                }                else {                    fail[next[now][i]]=next[fail[now]][i];                    Q.push(next[now][i]);                }        }    }}ac;int main(void){#ifdef LOCAL    freopen("/Users/zhaoyang/in.txt", "r", stdin);    //freopen("/Users/zhaoyang/out.txt", "w", stdout);#endif    ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);    int m, k;    while (~scanf("%d%d", &m, &k)) {        ac.init();        for (int i = 0; i < m; i++) {            scanf("%s", buf);            ac.insert(buf);        }        ac.build();        matrix A(ac.L);        for (int i = 0; i < ac.L; i++) {            if (!ac.end[i]) {                for (int j = 0; j < 4; j++) {                    int temp = ac.next[i][j];                    if (!ac.end[temp]) A.M[i][temp]++;                }            }        }        matrix B = A ^ k;        /*for (int i = 0; i < ac.L; i++) {            cout << i << ":  ";            for (int j = 0; j < ac.L; j++) {                cout << B.M[i][j] << " ";            }            cout << endl;        }*/        ll ans = 0;        for (int i = 0; i < ac.L; i++) ans += B.M[0][i] % mod;        printf("%lld\n", ans % mod);    }    return 0;}