poj 2778 DNA Sequence ac机+矩阵快速幂

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#矩阵快速幂 #ac机

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Language:

DNA Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total 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 3
AT
AC
AG
AA

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;
}

Language:

DNA Sequence