[日常摸鱼]poj2778 DNA Sequence

这题太神啦

题意：求长度为$n$的不包含给定DNA序列的DNA序列个数，给定的不超过10个

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构建出Trie图，用$danger[i]$来表示不能走到$i$，对于DNA序列结尾的结点$danger$设为1，构建$fail$指针的时候对于一个结点$i$的某个后缀如果$danger$为1那么$danger[i]$也应该为1.

然后根据Trie图再构造出对应的转移矩阵，自乘$n$次统计答案.

为什么自乘$n$次就行了？[感性理解.jpg]

人傻自带大常数？

hhh

#include<cstdio>
#include<cstring>
typedef long long lint;
const lint MOD=100000;
const int N=105;
struct matrix
{
    lint w[N][N];
    matrix(){memset(w,0,sizeof(w));}
};
int n,m,cnt,head,tail,ans;
int tr[N][5],fail[N],q[N],map[300];
bool danger[N];char s[N];
inline void insert(char *c)
{
    int len=strlen(c+1),k=0;
    for(register int i=1;i<=len;i++)
    {
        int t=map[(int)c[i]];
        if(!tr[k][t])tr[k][t]=++cnt;
        k=tr[k][t];
    }
    danger[k]=1;
}
inline void build()
{
    for(register int i=0;i<4;i++)if(tr[0][i])q[tail++]=tr[0][i],fail[tr[0][i]]=0;
    while(head<tail)
    {
        int k=q[head++];
        for(register int i=0;i<4;i++)
        {
            if(!tr[k][i])tr[k][i]=tr[fail[k]][i];
            else
            {
                fail[tr[k][i]]=tr[fail[k]][i];
                danger[tr[k][i]]|=danger[fail[tr[k][i]]];
                q[tail++]=tr[k][i];
            }
        }
    }
}
inline matrix mul(matrix a,matrix b)
{
    matrix res;
    for(register int i=0;i<=cnt;i++)
        for(register int k=0;k<=cnt;k++)if(a.w[i][k])
            for(register int j=0;j<=cnt;j++)
            {
                lint temp=(lint)(a.w[i][k]*b.w[k][j]);
                res.w[i][j]=(res.w[i][j]+temp);
                if(res.w[i][j]>MOD)res.w[i][j]%=MOD;
            }
    return res;
}
inline matrix powmod(matrix a,int b)
{
    matrix res;for(register int i=0;i<=cnt;i++)res.w[i][i]=1;
    for(;b;b>>=1,a=mul(a,a))if(b&1)res=mul(res,a);
    return res;
}
int main()
{
    //freopen("input.in","r",stdin);
    map['A']=0;map['C']=1;map['G']=2;map['T']=3;
    scanf("%d%d",&m,&n);
    for(register int i=1;i<=m;i++)
    {
        scanf("%s",s+1);insert(s);
    }
    build();matrix res;
    for(register int i=0;i<=cnt;i++)if(!danger[i])
    {
        for(register int j=0;j<4;j++)
            if(!danger[tr[i][j]])res.w[i][tr[i][j]]++;
    }
    res=powmod(res,n);
    for(register int i=0;i<=cnt;i++)
    {
        ans=ans+res.w[0][i];
        while(ans>MOD)ans-=MOD;
    }
    printf("%d",ans);
    return 0;
}