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
题意简单来说，就是【POJ1625】检查Censored!加了矩阵乘法优化
思路：
AC自动机+DP,处理失败指针（fail）时继承失败指针的标记（flag）
trie[v].flag|=trie[trie[trie[u].fail].ch[i]].flag;
o（n）太大，用矩阵乘法优化
注意：
结构体初始化清零，运算过程要mod，不然要开long long
代码：
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define mod 100000
using namespace std;
int m,n,cnt;
struct node{
long long e[105][105];
void clear(){memset(e,0,sizeof(e));}
}sequ,ans;
int rflag(char a)
{
if(a=='A') return 0;
if(a=='C') return 1;
if(a=='T') return 2;
if(a=='G') return 3;
}
struct fdfdfd{
int flag,fail,ch[5];
void clear(){flag=0; fail=0; memset(ch,0,sizeof(ch));}
}trie[105];
void insert(char a[],int len)
{
int root=0;
for(int i=0;i<len;++i)
{
int j=rflag(a[i]);
if(!trie[root].ch[j]) trie[cnt+1].clear(),trie[root].ch[j]=++cnt;
root=trie[root].ch[j];
}
trie[root].flag=1;
}
void getfail()
{
queue<int> q;
for(int i=0;i<4;++i)
if(trie[0].ch[i]) q.push(trie[0].ch[i]);
while(!q.empty())
{
int u=q.front(); q.pop();
for(int i=0;i<4;++i)
{
int v=trie[u].ch[i];
if(!v) trie[u].ch[i]=trie[trie[u].fail].ch[i];
else
{
trie[v].flag|=trie[trie[trie[u].fail].ch[i]].flag;
trie[v].fail=trie[trie[u].fail].ch[i];
q.push(v);
}
}
}
}
node Mul(node &a,node &b)
{
node temp; temp.clear();
for(int i=0;i<=cnt;++i)
for(int j=0;j<=cnt;++j)
{
temp.e[i][j]=0;
for(int k=0;k<=cnt;++k) temp.e[i][j]+=a.e[i][k]*b.e[k][j]%mod,temp.e[i][j]%=mod;
}
return temp;
}
int main()
{
scanf("%d%d",&m,&n);
trie[0].clear();
for(int i=1;i<=m;++i)
{
char temp[12]; scanf("%s",&temp);
insert(temp,strlen(temp));
}
getfail();
ans.clear(); sequ.clear();
for(int i=0;i<=cnt;++i)
for(int j=0;j<4;++j)
if(!trie[i].flag&&!trie[trie[i].ch[j]].flag) ++sequ.e[i][trie[i].ch[j]];
for(int i=0;i<=cnt;++i) ans.e[i][i]=1;
for(;n;n>>=1)
{
if(n&1) ans=Mul(ans,sequ);
sequ=Mul(sequ,sequ);
}
long long sum=0;
for(int i=0;i<=cnt;++i) sum=(sum+ans.e[0][i])%mod;
printf("%lld
",sum);
return 0;
}