poj 2778 DNA Sequence ac机+矩阵快速幂

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DNA Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11106 Accepted: 4243

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

Source

POJ Monthly--2006.03.26,dodo

题意：给定m个仅由A、C、G、T构成的非法串，问仅由A、C、G、T构成的长度为n的字符串共有多少个。

思路：先将m个非法串建ac机，用ac自动机求出初始矩阵，初始矩阵A.matrix[i][j]表示从自动机的节点i走一步不经过熟悉串的结尾有几种方法可以走到节点j,然后用对A.matrix矩阵进行二分快速幂求A^L，其中A.matrix[i][j]表示从i到j经过L步不经过熟悉串的结尾的方法数。对于初始矩阵，我们注意到对于自动机节点i来说，如果它的cnt!=0，那么它并没有存在的意义。所以我们对ac机上cnt=0的点进行重新编码,建立初始矩阵，最后答案为sum(A.matrix[0][k],0<=k<sz'),sz'为ac机上cnt=0的个数。需要注意matrix要用long long ,详见程序：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int MAXN=100+50;const int mod=100000;const int sigma_size=4;int n,m,head,tail,sz;ll l;struct node{    int cnt,id;    node *next[sigma_size],*fail;}trie[MAXN],*root,*que[MAXN];struct AC{    node *createnode()    {        for(int k=0;k<sigma_size;k++)            trie[sz].next[k]=NULL;        trie[sz].fail=NULL;        trie[sz].cnt=0,trie[sz].id=sz;        return &trie[sz++];    }    void init()    {        sz=0;        head=tail=0;        root=createnode();    }    int idx(char c)    {        if(c=='A') return 0;        if(c=='T') return 1;        if(c=='G') return 2;        return 3;    }    void insert(char *str)    {        node *p=root;        int len=strlen(str);        for(int i=0;i<len;i++)        {            int k=idx(str[i]);            if(p->next[k]==NULL)                p->next[k]=createnode();            p=p->next[k];        }        p->cnt++;    }    void get_fail()    {        que[tail++]=root;        while(head<tail)        {            node *p=que[head++];            for(int k=0;k<sigma_size;k++)                if(p->next[k])                {                    if(p==root)                        p->next[k]->fail=root;                    else                        p->next[k]->fail=p->fail->next[k];                    p->next[k]->cnt|=p->next[k]->fail->cnt;                    que[tail++]=p->next[k];                }                else                {                    if(p==root)                        p->next[k]=root;                    else                        p->next[k]=p->fail->next[k];                }        }    }}ac;struct Matrix{    ll matrix[MAXN][MAXN];}E;Matrix matrix_mul(Matrix a,Matrix b){    Matrix c;    for(int i=0;i<m;i++)        for(int j=0;j<m;j++)        {            c.matrix[i][j]=0;            for(int k=0;k<m;k++)                if(a.matrix[i][k] && b.matrix[k][j])                {                    c.matrix[i][j]+=a.matrix[i][k]*b.matrix[k][j];                    if(c.matrix[i][j]>=mod)                        c.matrix[i][j]%=mod;                }        }    return c;}Matrix matrix_pow(Matrix a,ll k){    Matrix c=E;    while(k)    {        if(k&1)            c=matrix_mul(c,a);        a=matrix_mul(a,a);        k>>=1;    }    return c;}int main(){    //freopen("text.txt","r",stdin);    for(int i=0;i<MAXN;i++)        E.matrix[i][i]=1;    while(~scanf("%d%I64d",&n,&l))    {        ac.init();        char str[15];        for(int i=0;i<n;i++)        {            scanf("%s",str);            ac.insert(str);        }        ac.get_fail();        Matrix A;        int u=0,v,num;        for(int i=0;i<sz;i++)            if(!trie[i].cnt)            {                v=0;                for(int j=0;j<sz;j++)                    if(!trie[j].cnt)                    {                        num=0;                        for(int k=0;k<sigma_size;k++)                            if(trie[i].next[k]->id==trie[j].id)                                num++;                        A.matrix[u][v]=num;                        v++;                    }               u++;            }        m=u;        A=matrix_pow(A,l);        ll ans=0;        for(int i=0;i<m;i++)        {            ans+=A.matrix[0][i];            if(ans>=mod)                ans%=mod;        }        printf("%I64d\n",ans);    }    return 0;}