POJ 2778 (AC自动机 矩阵快速幂)

题目链接:点击这里

题意:求不出现任意一个模式串的长度为m的文本串的数量.

建立AC自动机,文本串生成的过程可以看成是在自动机上某一个节点往4个方向上走一步,因为不能出现模式串,所以不能走到模式串结尾的节点.所以可以建立矩阵,a[i][j]表示从节点i到节点j的方案数. 根据邻接矩阵的定义, 这个矩阵的m次就是从某一个点走m步到达另一个点的方案数, 统计下就好了.

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>#include <queue>using namespace std;#define maxn 111#define mod 100000int n;long long m;char str[11][11];int tot;void change (char *s) {    int len = strlen (s);    for (int i = 0; i < len; i++) {        if (s[i] == 'A') s[i] = 'a';        else if (s[i] == 'C') s[i] = 'b';        else if (s[i] == 'T') s[i] = 'c';        else s[i] = 'd';    }    return ;}struct M {    long long a[maxn][maxn];    M () {        memset (a, 0, sizeof a);    }    M operator * (const M &gg) const {        M ans;        memset (ans.a, 0, sizeof ans.a);        for (int i = 0; i < tot; i++) {            for (int j = 0; j < tot; j++) {                for (int l = 0; l < tot; l++) {                    ans.a[i][j] += a[i][l]*gg.a[l][j];                    ans.a[i][j] %= mod;                }            }        }        return ans;    }    void show () {        for (int i = 0; i < tot; i++) {            for (int j = 0; j < tot; j++)                cout << a[i][j] << " ";            cout << endl;        }    }};M qpow (M a, long long k) {    M ans;    int i, j;    for (i = 0; i < tot; ++i)        for (j = 0; j < tot; ++j)            ans.a[i][j] = (i == j ? 1 : 0);    for(; k; k >>= 1) {        if (k&1) ans = ans*a;        a = a*a;    }    return ans;}struct trie {    int next[maxn][4], fail[maxn], end[maxn];    int root, cnt;    int new_node () {        memset (next[cnt], -1, sizeof next[cnt]);        end[cnt++] = 0;        return cnt-1;    }    void init () {        cnt = 0;        root = new_node ();    }    void insert (char *buf) {//字典树插入一个单词        int len = strlen (buf);        int now = root;        for (int i = 0; i < len; i++) {            int id = buf[i]-'a';            if (next[now][id] == -1) {                next[now][id] = new_node ();            }            now = next[now][id];        }        end[now]++;    }    void build () {//构建fail指针        queue <int> q;        fail[root] = root;        for (int i = 0; i < 4; i++) {            if (next[root][i] == -1) {                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 < 4; i++) {                if (next[now][i] == -1) {                    next[now][i] = next[fail[now]][i];                }                else {                    fail[next[now][i]] = next[fail[now]][i];                    q.push (next[now][i]);                }            }        }    }    int f[maxn];//安全节点对应的节点编号    long long query () {        M ans; memset (ans.a, 0, sizeof ans.a);        memset (f, -1, sizeof f);        tot = 0;        for (int i = 0; i < cnt; i++) if (!end[i]) {            bool flag = 1;            int tmp = i;            while (tmp != root) {                 if (end[tmp]) {                    flag = 0;                    break;                }                tmp = fail[tmp];             }            if (flag) {                f[i] = tot++;            }        }        for (int i = 0; i < cnt; i++) if (f[i] != -1) {//构造矩阵            int u = f[i];            for (int id = 0; id < 4; id++) {                int v = next[i][id];                if (f[v] != -1) {                    v = f[v];                    ans.a[u][v]++;                }            }        }        //ans.show ();        ans = qpow (ans, m);        long long res = 0;        for (int i = 0; i < tot; i++) {            res += ans.a[0][i];            res %= mod;        }        return res;    }}ac;int main () {    while (scanf ("%d%lld", &n, &m) == 2) {        ac.init ();        for (int i = 1; i <= n; i++) {            scanf ("%s", str[i]);            change (str[i]);            ac.insert (str[i]);        }        ac.build ();        long long ans = ac.query ();        printf ("%lld\n", ans);    }    return 0;}