poj 2778 AC自动机与矩阵连乘
来源：程序员人生 发布时间：2015-06-08 08:31:11 阅读次数：3163次
http://poj.org/problem?id=2778
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
/**
poj 2778 AC自动机与矩阵连乘
题目大意：给定1些模式串，问可以构造出多少中长度为n且不含模式串中的任何1个作为子串的字符串
解题思路：构造自动机，写出状态转移的矩阵，进行矩阵快速幂，其n次幂就表示长度为n。然后mat[0][i]就表示从根节点到状态点i长度为n的字符串有多少种
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
const int MOD=100000;
struct Matrix
{
int mat[110][110],n;
Matrix() {}
Matrix(int _n)
{
n=_n;
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
mat[i][j]=0;
}
}
}
Matrix operator *(const Matrix &b)const
{
Matrix ret=Matrix(n);
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
for(int k=0; k<n; k++)
{
int tmp=(long long)mat[i][k]*b.mat[k][j]%MOD;
ret.mat[i][j]=(ret.mat[i][j]+tmp)%MOD;
}
}
}
return ret;
}
};
struct Trie
{
int next[110][4],fail[110],end[110];
int root,L;
int newnode()
{
for(int i=0; i<4; i++)
next[L][i]=⑴;
end[L++]=0;
return L⑴;
}
void init()
{
L=0;
root=newnode();
}
int getch(char ch)
{
if(ch=='A')
return 0;
if(ch=='C')
return 1;
if(ch=='G')
return 2;
return 3;
}
void insert(char *s)
{
int len=strlen(s);
int now=root;
for(int i=0; i<len; i++)
{
if(next[now][getch(s[i])]==⑴)
next[now][getch(s[i])]=newnode();
now=next[now][getch(s[i])];
}
end[now]=1;
}
void build()
{
queue <int>Q;
for(int i=0; i<4; i++)
{
if(next[root][i]==⑴)
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]]==1)
end[now]=1;
for(int i=0; i<4; i++)
{
if(next[now][i]==⑴)
next[now][i]=next[fail[now]][i];
else
{
fail[next[now][i]]=next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
}
Matrix getMatrix()
{
Matrix res=Matrix(L);
for(int i=0; i<L; i++)
{
for(int j=0; j<4; j++)
{
if(end[next[i][j]]==0)
res.mat[i][next[i][j]]++;
}
}
return res;
}
} ac;
char buf[20];
Matrix pow_M(Matrix a,int n)
{
Matrix ret=Matrix(a.n);
for(int i=0; i<ret.n; i++)
{
ret.mat[i][i]=1;
}
Matrix tmp=a;
while(n)
{
if(n&1)ret=ret*tmp;
tmp=tmp*tmp;
n>>=1;
}
return ret;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
ac.init();
for(int i=0;i<n;i++)
{
scanf("%s",buf);
ac.insert(buf);
}
ac.build();
Matrix a=ac.getMatrix();
a=pow_M(a,m);
int ans=0;
for(int i=0;i<a.n;i++)
{
ans=(ans+a.mat[0][i])%MOD;
}
printf("%d
",ans);
}
return 0;
}
生活不易，码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
