Poj 2778 [AC自动机，矩阵乘法]

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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.

题意：有m种DNA序列是有疾病的，问有多少种长度为n的DNA序列不包含任何一种有疾病的DNA序列。（仅含A,T,C,G四个字符）
dp[i][j]=sum(dp[i-1][k])
枚举第i位填了’A’,’C’,’G’,’T’中的哪一个，自动机从k走到了哪里。
把i看作行，j看做列，那么dp[i][]和dp[i-1][]之间的关系显然和dp[i+1][]和dp[i-1][]的关系相同。
因此，可以用构造转移矩阵方法优化。。。矩阵就是dp【i][j]表示从I到j结点走一步（加一个字母）有几种方法，注意不能走到任何模式串的尾巴也不能它来转移，所以相当于尾结点的行列都不要，然后n次方，就是加了n个字母（题目要求了长度），初始化时候，只把0 0设置为1（没有字母），最后所有（0，i）的值加起来，就是0号结点转移到I的。
注意m=0特判。。。

#include <cstdio>#include <iostream>#include <cstring>#include <queue>using namespace std;int n,m,node;const int maxn = 15;const long long MOD = 100000;char ss[maxn][maxn];int val[maxn*10],last[maxn*10],ch[maxn*10][maxn],fail[maxn*10],kmax;int getpos(char a){    if(a == 'A')    return 0;    if(a == 'T')    return 1;    if(a == 'C')    return 2;    if(a == 'G')    return 3;}void Insert(int v){    int len = strlen(ss[v]);    int tmp = 0;    for(int i = 0; i < len ; i++){        int pos = getpos(ss[v][i]);        if(ch[tmp][pos] == 0){            memset(ch[node],0,sizeof(ch[node]));            val[node] = 0;            ch[tmp][pos] = node++;        }        tmp = ch[tmp][pos];    }    val[tmp] = v;}void getfail(){    queue<int> q;    int u = 0;    for(int i = 0; i < 4; i++){        if(ch[u][i]){            int tmp = ch[u][i];            fail[tmp] = 0;            q.push(tmp);            last[tmp] = 0;        }    }    while(!q.empty()){        int tmp = q.front();        q.pop();        for(int i = 0; i < 4; i++){            if(ch[tmp][i]){                int v = fail[tmp];                while(v && ch[v][i]==0) v = fail[v];                fail[ch[tmp][i]] = ch[v][i];                q.push(ch[tmp][i]);               // if(ch[tmp][i]==2)   printf("asd   ch[v][i] = %d \n",ch[v][i]);               last[ch[tmp][i]] = val[ch[v][i]]?ch[v][i] : last[ch[v][i]];            }        }    }}void init(){    node = 1;    kmax = 0;    memset(ch[0],0,sizeof(ch[0]));    for(int i = 1; i <= m ; i++){        scanf("%s",ss[i]);        Insert(i);    }}struct Matrix{    long long m[maxn*10][maxn*10];}dp,ini;void print(Matrix a){    for(int i = 0 ; i < 15 ; i++){        for(int j = 0 ; j < 15 ; j++)            printf("%lld ",a.m[i][j]);        printf("\n");    }    cout <<"\n";}void make_matrix(){   // cout <<"node = " << node <<endl;    memset(dp.m,0,sizeof(dp.m));    for(int i = 0 ; i < node; i++){        if(val[i]||val[last[i]])  continue;        for(int j = 0 ; j < 4; j++){            //printf("i = %d  j = %d \n",i,j);            //printf("last[%d]  = %d    val = %d\n",ch[i][j],last[ch[i][j]],val[ch[i][j]]);            if(last[ch[i][j]]||val[ch[i][j]])  continue;            int v = i;            if(ch[v][j]!=0){                dp.m[i][ch[v][j]] += 1;                kmax = max(kmax,max(i,ch[v][j]));            }            else{                while(v && ch[v][j] == 0 )    v = fail[v];                if(last[ch[v][j]]||val[ch[v][j]])  continue;                dp.m[i][ch[v][j]] += 1;                kmax = max(kmax,max(i,ch[v][j]));            }        }    }   // print(dp);}Matrix Mul(Matrix a,Matrix b){    Matrix p;    for(int i = 0 ;i <= kmax ; i++)        for(int j = 0 ; j <= kmax; j++)            p.m[i][j] = 0;    for(int i =  0 ;i <= kmax ;i++){        for(int j = 0 ; j <= kmax ; j++){                for(int k = 0 ; k <= kmax ; k++)                    p.m[i][j] = (p.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;        }    }    return p;}Matrix pow(Matrix a, int n){    Matrix p;     for(int i = 0 ;i <= kmax ; i++)        for(int j = 0 ; j <= kmax; j++)            p.m[i][j] = 0;    for(int i = 0 ; i <= kmax ; i++)  p.m[i][i] = 1;    while(n){        if(n & 1)   p = Mul(p,a);        n >>= 1;        a = Mul(a,a);    }   // print(p);    return p;}void sov(){     for(int i = 0 ;i <= kmax ; i++)        for(int j = 0 ; j <= kmax; j++)            ini.m[i][j] = 0;    ini.m[0][0] = 1;    //cout <<"n = "<<n<<endl;    ini =  Mul(ini,pow(dp,n));    int ans = 0;    for(int i = 0 ; i < node; i++)        ans = (ans + ini.m[0][i])%MOD;    printf("%d\n",ans);}/*3 10ATGATCTGAC*/int qpow(int n){    int a = 1,p = 4;    while(n){        if(n & 1)   a = (a*p)%MOD;        n >>= 1;        p = (p*p)%MOD;    }    return a;}int main(){    while(~scanf("%d%d",&m,&n)){        if(m == 0){           // cout <<"Asd"<<endl;            printf("%d\n",qpow(n));            continue;        }        init();        getfail();        make_matrix();        sov();    }    return 0;}

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