POJ 2778 DNA Sequence （AC自动机+矩阵加速，4级）

E - DNA SequenceCrawling in process...Crawling failedTime Limit:1000MSMemory Limit:65536KB 64bit IO Format:%I64d & %I64uSubmitStatusAppoint description:System Crawler (2013-05-30)DescriptionIt'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 animalmay 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.InputFirst 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.OutputAn integer, the number of DNA sequences, mod 100000.Sample Input

4 3
AT
AC
AG
AA

Sample Output

36

思路：直接用AC自动机预处理出失败指针转移态，然后根据失败指针的转移态建立递推矩阵，外加个快速幂轻松搞定。

mod很占时间能不mod尽量少mod

#include<iostream>#include<cstring>#include<cstdio>#include<queue>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))#define LL unsigned long longusing namespace std;const int msize=510;const int sig=4;const int mod=100000;class Matrix{public:LL f[140][140];int n;Matrix();Matrix(int x){n=x;FOR(i,0,x-1)FOR(j,0,x-1)f[i][j]=0;}Matrix operator*(const Matrix&b)const{Matrix c=Matrix(n);FOR(i,0,n-1)FOR(j,0,n-1){FOR(k,0,n-1)c.f[i][j]+=f[i][k]*b.f[k][j];c.f[i][j]%=mod;}return c;}};class AC_Machine{public:int f[msize],ch[msize][sig],sz;bool val[msize];void clear(){sz=1;clr(ch[0],0);val[0]=0;}int idx(char x){if(x=='A')return 0;if(x=='G')return 1;if(x=='C')return 2;if(x=='T')return 3;}void insert(char*s){int u=0,v,c;for(int i=0;s[i];++i){c=idx(s[i]);if(!ch[u][c]){clr(ch[sz],0);val[sz]=0;ch[u][c]=sz++;}u=ch[u][c];}val[u]=1;}void getFail(){int u=0,v,r;queue<int>Q;FOR(c,0,sig-1){u=ch[0][c];if(u){f[u]=0;Q.push(u);}}while(!Q.empty()){r=Q.front();Q.pop();val[r]|=val[ f[r] ];FOR(c,0,sig-1){u=ch[r][c];if(!u){ch[r][c]=ch[ f[r] ][c];continue;}v=f[r];Q.push(u);while(v&&!ch[v][c])v=f[v];f[u]=ch[v][c];}}}Matrix getMatrix(){Matrix ret=Matrix(sz);FOR(i,0,sz-1)FOR(j,0,3)if(!val[ch[i][j]]){ret.f[i][ ch[i][j] ]++;}return ret;}};Matrix M_pow(Matrix a,int m){Matrix ret=Matrix(a.n);FOR(i,0,a.n-1)ret.f[i][i]=1;while(m){if(m&1)ret=ret*a;a=a*a;m>>=1;}return ret;}AC_Machine ac;char s[55];int n,m;int main(){while(~scanf("%d%d",&n,&m)){ac.clear();FOR(i,1,n){scanf("%s",s);ac.insert(s);}ac.getFail();Matrix ret=ac.getMatrix();ret=M_pow(ret,m);LL ans=0;FOR(i,0,ac.sz-1)ans+=ret.f[0][i];ans%=mod;printf("%lld\n",ans);}}

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