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DNA Sequence
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 |
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非常经典的AC自动机+矩阵。。。。。
节点比较少，需要构造的串很长。。。所以直接把各种转移状态记录到矩阵里，用快速幂解决。。。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=10000;
const int MOD=100000;
int idx(char x)
{
if(x==‘A‘) return 0;
else if(x==‘T‘) return 1;
else if(x==‘C‘) return 2;
else if(x==‘G‘) return 3;
}
int ch[maxn][5],fail[maxn],end[maxn];
int root,sz;
char str[200];
int newnode()
{
memset(ch[sz],-1,sizeof(ch[sz]));
end[sz++]=0;
return sz-1;
}
void ac_init()
{
sz=0; root=newnode();
}
void ac_insert(char str[])
{
int len=strlen(str);
int now=root;
for(int i=0;i<len;i++)
{
if(ch[now][idx(str[i])]==-1)
ch[now][idx(str[i])]=newnode();
now=ch[now][idx(str[i])];
}
end[now]++;
}
void ac_build()
{
queue<int> q;
fail[root]=root;
for(int i=0;i<4;i++)
{
int& temp=ch[root][i];
if(temp==-1)
{
ch[root][i]=root;
}
else
{
fail[ch[root][i]]=root;
q.push(ch[root][i]);
}
}
while(!q.empty())
{
int now=q.front(); q.pop();
end[now]+=end[fail[now]];
for(int i=0;i<4;i++)
{
if(ch[now][i]==-1)
{
ch[now][i]=ch[fail[now]][i];
}
else
{
fail[ch[now][i]]=ch[fail[now]][i];
q.push(ch[now][i]);
}
}
}
}
int n,m;
struct MARTRIX
{
int _x,_y;
long long int martrix[200][200];
MARTRIX () {}
MARTRIX (int n)
{
_x=_y=n;
memset(martrix,0,sizeof(martrix));
}
void getONE(int n)
{
for(int i=0;i<n;i++)
{
martrix[i][i]=1;
}
}
MARTRIX operator * (const MARTRIX& b) const
{
int n=b._x;
MARTRIX ret(n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
int temp=0;
for(int k=0;k<n;k++)
{
temp=(temp+martrix[i][k]*b.martrix[k][j])%MOD;
}
ret.martrix[i][j]=temp%MOD;
}
}
return ret;
}
};
MARTRIX QuickPOW(MARTRIX M,int n)
{
MARTRIX ans;
ans.getONE(M._x);
while(n)
{
if(n&1)
{
ans=ans*M;
}
M=M*M;
n>>=1;
}
return ans;
}
int main()
{
while(scanf("%d%d",&m,&n)!=EOF)
{
ac_init();
for(int i=0;i<m;i++)
{
scanf("%s",str);
ac_insert(str);
}
ac_build();
MARTRIX mat(sz);
for(int i=0;i<sz;i++)
{
if(end[i]) continue;
for(int j=0;j<4;j++)
{
int p=ch[i][j];
if(end[p]||end[fail[p]]) continue;
mat.martrix[i][p]++;
}
}
MARTRIX ret=QuickPOW(mat,n);
long long int ans=0;
for(int i=0;i<sz;i++)
{
ans=(ans+ret.martrix[0][i])%MOD;
}
printf("%I64d\n",ans);
}
return 0;
}
POJ 2778 DNA Sequence,布布扣,bubuko.com
原文：http://blog.csdn.net/ck_boss/article/details/23716325