poj 2778 DNA Sequence //AC自动机+矩阵乘法

DNA Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5080 Accepted: 1766

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 3ATACAGAA

Sample Output

36

Source

POJ Monthly--2006.03.26,dodo

 

 

/*少用取模，那东西太耗时间了*/#include<cstdio>#include<cstring>const int NODE=13*13;const int CH=4;int tree[NODE][CH],cnt;int word[NODE];int fail[NODE];int Que[NODE];int num[150];int m,n;int matnum,matcal[NODE];__int64 mm[NODE][NODE],r[NODE][NODE],tem[NODE][NODE];void Ins(char *a,int val){ int p=0; for(; *a; a++) { int c=num[*a]; if(!tree[p][c]) { memset(tree[cnt],0,sizeof(tree[cnt]));//初始化新的节点 word[cnt]=0; tree[p][c]=cnt++; } p=tree[p][c]; } word[p]=val;//做一个标记，说明这是病毒串}void AC(){ int *s=Que,*e=Que; for(int i=0; i<CH; ++i) if(tree[0][i]) { fail[tree[0][i]]=0;//修改失败指针 *e++=tree[0][i]; } while(s!=e) { int p=*s++; for(int i=0; i<CH; ++i) { if(tree[p][i]) { int v=tree[p][i]; *e++=v; fail[v]=tree[fail[p]][i]; word[v]|=word[fail[v]];//Wa了N次 } else { tree[p][i]=tree[fail[p]][i]; } } }}char s[13];void Input(){ fail[0]=0; memset(tree[0],0,sizeof(tree[0]));//初始化根节点，开始都指向根节点 //从1开始，0有特殊使用 cnt=1;//已经有根节点,节点是从1<=i<cnt num['A']=0; num['T']=1; num['C']=2; num['G']=3; for(int i=1; i<=m; i++) { scanf("%s",s); Ins(s,1); }}void init()//建立矩阵的关系{ memset(mm,0,sizeof(mm)); memset(matcal,0,sizeof(matcal)); matnum=0;//矩阵从0<=i<matnum for(int i=0; i<cnt; ++i) if(word[i]==0) matcal[i]=matnum++; for(int i=0; i<cnt; ++i) if(word[i]==0) { for(int j=0; j<CH; ++j) { if(word[tree[i][j]]==0)//不是病毒串 { mm[matcal[i]][matcal[tree[i][j]]]++; } } }}void Mul(__int64 a[NODE][NODE],__int64 b[NODE][NODE]){ memset(tem,0,sizeof(tem)); for(int i=0; i<matnum; ++i) for(int j=0; j<matnum; ++j) { for(int k=0; k<matnum; ++k) tem[i][j]=tem[i][j]+a[i][k]*b[k][j]; tem[i][j]=tem[i][j]%100000; }}void work(){ for(int i=0; i<matnum; ++i) r[i][i]=1; while(n) { if(n&1) { Mul(mm,r); memcpy(r,tem,sizeof(tem)); } n>>=1; Mul(mm,mm); memcpy(mm,tem,sizeof(tem)); } int res=0; for(int i=0; i<matnum; ++i) res+=r[0][i]; res=res%100000; printf("%d/n",res);}int main(){ scanf("%d%d",&m,&n); Input(); AC(); init(); /*for(int i=0;i<matnum;++i) { for(int j=0;j<matnum;++j) { printf("%d ",mm[i][j]); } printf("/n"); }*/ work(); return 0;}