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**:
***********************1599664856@qq.com**********************/

#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;
}