HDU 2243 postgraduate entrance-word complex (AC automation + matrix)

Hi, Chinese question... I won't talk about it.

First, poj 2778 knows that the AC automatic mechanism can be used to find the number of non-virus strings in strings with a length of L.

The general idea of poj 2778 is to first create an AC automatic machine with all the virus strings, and then connect all non-word nodes on the AC automatic machine to an edge.

In Discrete Mathematics, if [I] [J] in matrix A represents the number of methods that an I node can walk to a j node through an edge.

Then the [I] [J] of the matrix A * A indicates the number of methods that an I node can go to a j node through two edges.

Now that we know this method, we have a clear requirement.

Ans = 26 + 26 ^ 2 + 26 ^ 3 + .... + 26 ^ L-the number of non-virus strings with a length of 1-the number of non-virus strings with a length of 2 -... -The length is the number of non-virus strings.

Solve two problems

Modulo 2 ^ 64. Knowing that 2 ^ 64 is the maximum value of long, we can directly open it into unsigned long... and then rest assured that overflow is the modulus.

We know that the matrix ^ L requires all the sums, and we need to use a matrix in the matrix. That is, the sum of matrices.

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#define N 35using namespace std;typedef unsigned long long ll;const char tab = 'a';const int max_next = 26;struct trie{    struct trie *fail;    struct trie *next[max_next];    int isword;    int index;};struct AC{    trie *que[100005],*root,ac[100005];    int head,tail;    int idx;    trie *New()    {        trie *temp=&ac[idx];        for(int i=0;i<max_next;i++)temp->next[i]=NULL;        temp->fail=NULL;        temp->isword=0;        temp->index=idx++;        return temp;    }    void init()    {        idx=0;        root=New();    }    void Insert(trie *root,char *word,int len){        trie *t=root;        for(int i=0;i<len;i++){            if(t->next[word[i]-tab]==NULL)                t->next[word[i]-tab]=New();            t=t->next[word[i]-tab];        }        t->isword++;    }    void acbuild(trie *root){        int head=0,tail=0;        que[tail++]=root;        root->fail=NULL;        while(head<tail){            trie *temp=que[head++],*p;            for(int i=0;i<max_next;i++){                 if(temp->next[i]){                    if(temp==root)temp->next[i]->fail=root;                    else {                        p=temp->fail;                        while(p!=NULL){                            if(p->next[i]){                                temp->next[i]->fail=p->next[i];                                break;                            }                            p=p->fail;                        }                        if(p==NULL)temp->next[i]->fail=root;                    }                    if(temp->next[i]->fail->isword)temp->next[i]->isword++;                    que[tail++]=temp->next[i];                 }                 else if(temp==root)temp->next[i]=root;                 else temp->next[i]=temp->fail->next[i];            }        }    }    void tra()    {        for(int i=0;i<idx;i++)        {            if(ac[i].fail!=NULL)printf("fail = %d ",ac[i].fail->index);            for(int k=0;k<max_next;k++)                printf("%d ",ac[i].next[k]->index);            puts("");        }    }}sa;struct matrix{    int r,c;    ll data[N][N];    matrix(){}    matrix(int _r,int _c):r(_r),c(_c){memset(data,0,sizeof data);}    friend matrix operator * (const matrix A,const matrix B)    {        matrix res;        res.r=A.r;res.c=B.c;        memset(res.data,0,sizeof res.data);        for(int i=0;i<A.r;i++)        {            for(int j=0;j<B.c;j++)            {                for(int k=0;k<A.c;k++)                {                    if(A.data[i][k] && B.data[k][j]){                        res.data[i][j]+=A.data[i][k]*B.data[k][j];                        //res.data[i][j]%=mod;                    }                }            }        }        return res;    }    friend matrix operator + (const matrix A,const matrix B)    {        matrix res;        res.r=A.r;res.c=A.c;        memset(res.data,0,sizeof res.data);        for(int i=0;i<A.r;i++)        {            for(int j=0;j<A.c;j++)            {                res.data[i][j]=A.data[i][j]+B.data[i][j];                //res.data[i][j]%=mod;            }        }        return res;    }    friend matrix operator - (const matrix A,const matrix B)    {        matrix res;        res.r=A.r;res.c=A.c;        memset(res.data,0,sizeof res.data);        for(int i=0;i<A.r;i++)        {            for(int j=0;j<A.c;j++)            {                res.data[i][j]=A.data[i][j]-B.data[i][j];                //res.data[i][j]=(res.data[i][j]%mod+mod)%mod;            }        }        return res;    }    friend matrix operator ^ (matrix A,int n)    {        matrix res;        res.r=A.r;res.c=A.c;        memset(res.data,0,sizeof res.data);        for(int i=0;i<A.r;i++)res.data[i][i]=1;        while(n)        {            if(n&1)res=res*A;            A=A*A;            n>>=1;        }        return res;    }    void print()    {        for(int i=0;i<r;i++)        {            for(int j=0;j<c;j++)                printf("%d ",data[i][j]);            puts("");        }    }}E,zero;char word[10];struct supermatrix{    matrix ret[2][2];    friend supermatrix operator * (const supermatrix A,const supermatrix B)    {        supermatrix res;        for(int i=0;i<2;i++)            for(int j=0;j<2;j++)res.ret[i][j]=zero;        for(int i=0;i<2;i++)        {            for(int j=0;j<2;j++)            {                for(int k=0;k<2;k++)                {                    res.ret[i][j]=res.ret[i][j]+A.ret[i][k]*B.ret[k][j];                }            }        }        return res;    }    friend supermatrix operator + (const supermatrix A,const supermatrix B)    {        supermatrix res;        for(int i=0;i<2;i++)for(int j=0;j<2;j++)res.ret[i][j]=zero;        for(int i=0;i<2;i++)        {            for(int j=0;j<2;j++)            {                res.ret[i][j]=A.ret[i][j]+B.ret[i][j];            }        }        return res;    }    friend supermatrix operator ^ (supermatrix A,ll n)    {        supermatrix res;        for(int i=0;i<2;i++)for(int j=0;j<2;j++)res.ret[i][j]=zero;        for(int i=0;i<2;i++)res.ret[i][i]=E;        while(n)        {            if(n&1)res=res*A;            A=A*A;            n>>=1;        }        return res;    }};int main(){    int n,L;    while(scanf("%d%d",&n,&L)!=EOF)    {        sa.init();        for(int i=1;i<=n;i++)        {            scanf("%s",word);            sa.Insert(sa.root,word,strlen(word));        }        sa.acbuild(sa.root);        E=matrix(sa.idx,sa.idx);        for(int i=0;i<N;i++)E.data[i][i]=1;        zero=matrix(sa.idx,sa.idx);        matrix origin=matrix(sa.idx,sa.idx);        for(int i=0;i<sa.idx;i++)        {            if(sa.ac[i].isword==0)            {                for(int d=0;d<max_next;d++)                {                    int temp=sa.ac[i].next[d]->index;                    if(sa.ac[i].next[d]->isword==0)                    {                        origin.data[i][temp]++;                    }                }            }        }        supermatrix A;        A.ret[0][0]=A.ret[0][1]=E;        A.ret[1][0]=zero;        A.ret[1][1]=origin;        A=A^L;        supermatrix f;        f.ret[0][0]=f.ret[0][1]=f.ret[1][1]=zero;        f.ret[1][0]=origin;        matrix fans=A.ret[0][0]*zero+A.ret[0][1]*origin;        ll ans=0;        for(int i=0;i<sa.idx;i++)        {            if(sa.ac[i].isword==0)            {                ans+=fans.data[0][i];            }        }        matrix I=matrix(2,2);        I.data[0][0]=I.data[0][1]=1;        I.data[1][1]=26;        I.data[1][0]=0;        I=I^L;        printf("%I64u\n",I.data[0][1]*26-ans);    }    return 0;}

HDU 2243 postgraduate entrance-word complex (AC automation + matrix)