DNA Sequence
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12252 | Accepted: 4661 |
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
给出患病的DNA序列，问序列长度为n的，且不包含患病的DNA序列有多少种。
首先处理患病的DNA串，连接为字典树后，添加fail指针，完成AC自动机，给每个节点一个编号，然后用矩阵统计每个节点走一步可以走到的节点的种数，其中，不能包含会患病的序列，最后矩阵相乘。
注意1.__int64 相乘会超出范围。
注意2.处理自动机时，注意，如果发现某个节点的fail会返回到一个代表序列结束的节点上，那么这个节点也是不可达的。不能被统计到矩阵中。
注意3.矩阵快速幂要写成非递归的形式。
给出测试案例：
2 1
ACG
C
其中矩阵应该为
2 1 0 0 0
2 1 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std ;
#define MOD 100000
#define LL __int64
struct node{
int flag , id ;
node *next[4] , *fail ;
};
struct nnode{
LL Map[110][110] , n ;
};
queue <node*> que ;
char c[5] = "ACGT" ;
char str[20] ;
int num , vis[110] ;
node *newnode()
{
node *p = new node ;
p->flag = 0 ;
p->id = num++ ;
p->fail = NULL ;
for(int i = 0 ; i < 4 ; i++)
p->next[i] = NULL ;
return p ;
}
void settree(char *s,node *rt,int temp)
{
int i , k , l = strlen(s) ;
node *p = rt ;
for(i = 0 ; i < l ; i++)
{
for(k = 0 ; k < 4 ; k++)
if( s[i] == c[k] )
break ;
if( p->next[k] == NULL )
p->next[k] = newnode() ;
p = p->next[k] ;
}
p->flag = 1 ;
return ;
}
void setfail(node *rt)
{
int i ;
node *p = rt , *temp ;
p->fail = NULL ;
while( !que.empty() ) que.pop() ;
que.push(p) ;
while( !que.empty() )
{
p = que.front() ;
que.pop() ;
for(i = 0 ; i < 4 ; i++)
{
if( p->next[i] )
{
temp = p->fail ;
while( temp && !temp->next[i] )
temp = temp->fail ;
p->next[i]->fail = temp ? temp->next[i] : rt ;
if( temp != NULL && temp->next[i]->flag )
p->next[i]->flag = 1 ;
que.push(p->next[i]) ;
}
else
p->next[i] = p == rt ? rt : p->fail->next[i] ;
}
}
}
nnode setmat(node *rt)
{
int i , j , u , v ;
nnode q ;
node *p = rt ;
while( !que.empty() ) que.pop() ;
memset(q.Map,0,sizeof(q.Map)) ;
memset(vis,0,sizeof(vis)) ;
que.push(p) ;
q.n = num ;
vis[ p->id ] = 1 ;
while( !que.empty() )
{
p = que.front() ;
que.pop() ;
u = p->id ;
for(i = 0 ; i < 4 ; i++)
{
if( !p->flag && !p->next[i]->flag )
q.Map[ p->id ][ p->next[i]->id ]++ ;
if( !vis[p->next[i]->id] )
{
vis[ p->next[i]->id ] = 1 ;
que.push( p->next[i] ) ;
}
}
}
return q ;
}
nnode mul(nnode a,nnode b)
{
nnode c ;
c.n = a.n ;
int i , j , k ;
for(i = 0 ; i < a.n ; i++)
{
for(j = 0 ; j < a.n ; j++)
{
c.Map[i][j] = 0 ;
for(k = 0 ; k < a.n ; k++)
c.Map[i][j] = ( c.Map[i][j] + a.Map[i][k]*b.Map[k][j] ) % MOD ;
}
}
return c ;
}
nnode pow(nnode p,int k)
{
nnode temp ;
int i , j ;
temp.n = p.n ;
memset(temp.Map,0,sizeof(temp.Map)) ;
for(i = 0 ; i < p.n ; i++)
temp.Map[i][i] = 1 ;
while( k )
{
if( k&1 )
temp = mul(temp,p) ;
p = mul(p,p) ;
k >>= 1 ;
}
return temp ;
}
int main()
{
int n , m , i , j ;
node *rt ;
nnode p ;
while( scanf("%d %d", &m, &n) != EOF )
{
num = 0 ;
rt = newnode() ;
for(i = 1 ; i <= m ; i++)
{
scanf("%s", str) ;
settree(str,rt,i) ;
}
setfail(rt) ;
p = setmat(rt) ;
/*for(i = 0 ; i < p.n ; i++)
{
for(j = 0 ; j < p.n ; j++)
printf("%d ", p.Map[i][j]) ;
printf("\n") ;
}*/
p = pow(p,n) ;
LL ans = 0 ;
for(i = 0 ; i < p.n ; i++)
ans = ( ans + p.Map[0][i] ) % MOD ;
printf("%d\n", ans) ;
}
return 0 ;
}