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 100000

int 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;
}