DNA Sequence
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9899 | Accepted: 3717 |
Description
It's well known that DNA Sequence is a sequence only contains A, C, T and G, and it's very useful to analyze a segment of DNA Sequence，For example, if a animal's DNA sequence contains segment ATC then it may mean that the animal may have a genetic disease. Until now scientists have found several those segments, the problem is how many kinds of DNA sequences of a species don't contain those segments.
Suppose that DNA sequences of a species is a sequence that consist of A, C, T and G，and the length of sequences is a given integer n.
Input
First line contains two integer m (0 <= m <= 10), n (1 <= n <=2000000000). Here, m is the number of genetic disease segment, and n is the length of sequences.
Next m lines each line contain a DNA genetic disease segment, and length of these segments is not larger than 10.
Output
An integer, the number of DNA sequences, mod 100000.
Sample Input
4 3
AT
AC
AG
AA
Sample Output
36
Source
A自动机。
要求长度为n，不包含病毒串的个数。
首先利用AC自动机实现状态的转移。
AC自动机其实就和状态机类似的，可以产生L个状态。
然后根据状态间能不能转移，构造一个矩阵。
最后矩阵快速幂求解
//============================================================================
// Name : HDU.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================ #include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
struct Matrix
{
unsigned long long mat[][];
int n;
Matrix(){}
Matrix(int _n)
{
n=_n;
for(int i=;i<n;i++)
for(int j=;j<n;j++)
mat[i][j] = ;
}
Matrix operator *(const Matrix &b)const
{
Matrix ret = Matrix(n);
for(int i=;i<n;i++)
for(int j=;j<n;j++)
for(int k=;k<n;k++)
ret.mat[i][j]+=mat[i][k]*b.mat[k][j];
return ret;
}
};
unsigned long long pow_m(unsigned long long a,int n)
{
unsigned long long ret=;
unsigned long long tmp = a;
while(n)
{
if(n&)ret*=tmp;
tmp*=tmp;
n>>=;
}
return ret;
}
Matrix pow_M(Matrix a,int n)
{
Matrix ret = Matrix(a.n);
for(int i=;i<a.n;i++)
ret.mat[i][i] = ;
Matrix tmp = a;
while(n)
{
if(n&)ret=ret*tmp;
tmp=tmp*tmp;
n>>=;
}
return ret;
}
struct Trie
{
int next[][],fail[];
bool end[];
int root,L;
int newnode()
{
for(int i = ;i < ;i++)
next[L][i] = -;
end[L++] = false;
return L-;
}
void init()
{
L = ;
root = newnode();
}
void insert(char buf[])
{
int len = strlen(buf);
int now = root;
for(int i = ;i < len;i++)
{
if(next[now][buf[i]-'a'] == -)
next[now][buf[i]-'a'] = newnode();
now = next[now][buf[i]-'a'];
}
end[now] = true;
}
void build()
{
queue<int>Q;
fail[root]=root;
for(int i = ;i < ;i++)
if(next[root][i] == -)
next[root][i] = root;
else
{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
while(!Q.empty())
{
int now = Q.front();
Q.pop();
if(end[fail[now]])end[now]=true;
for(int i = ;i < ;i++)
if(next[now][i] == -)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
Matrix getMatrix()
{
Matrix ret = Matrix(L+);
for(int i = ;i < L;i++)
for(int j = ;j < ;j++)
if(end[next[i][j]]==false)
ret.mat[i][next[i][j]] ++;
for(int i = ;i < L+;i++)
ret.mat[i][L] = ;
return ret;
}
void debug()
{
for(int i = ;i < L;i++)
{
printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
for(int j = ;j < ;j++)
printf("%2d",next[i][j]);
printf("]\n");
}
}
};
char buf[];
Trie ac;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n,L;
while(scanf("%d%d",&n,&L)==)
{
ac.init();
for(int i = ;i < n;i++)
{
scanf("%s",buf);
ac.insert(buf);
}
ac.build();
Matrix a = ac.getMatrix();
a = pow_M(a,L);
unsigned long long res = ;
for(int i = ;i < a.n;i++)
res += a.mat[][i];
res--; /*
* f[n]=1 + 26^1 + 26^2 +...26^n
* f[n]=26*f[n-1]+1
* {f[n] 1} = {f[n-1] 1}[26 0;1 1]
* 数是f[L]-1;
* 此题的L<2^31.矩阵的幂不能是L+1次，否则就超时了
*/
a = Matrix();
a.mat[][]=;
a.mat[][] = a.mat[][] = ;
a=pow_M(a,L);
unsigned long long ans=a.mat[][]+a.mat[][];
ans--;
ans-=res;
cout<<ans<<endl;
}
return ;
}