hdu 2243 考研路茫茫——单词情结 //AC自动机

 

考研路茫茫——单词情结

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 520    Accepted Submission(s): 143

#include<cstdio>#include<cstring>typedef unsigned __int64 ull;const int NODE=35;const int CH=26;int tree[NODE][CH],cnt;int word[NODE];int fail[NODE];int Que[NODE];int num[150];int matnum,matcal[NODE];ull mm[NODE][NODE],tem[NODE][NODE],mat[NODE][NODE];void Ins(char *a,int val){ int p=0; for(; *a; a++) { int c=num[*a]; if(!tree[p][c]) { memset(tree[cnt],0,sizeof(tree[cnt]));//初始化新的节点 fail[cnt]=0;//更新 word[cnt]=0; tree[p][c]=cnt++; } p=tree[p][c]; } word[p]=val;//做一个标记，说明这是病毒串}void AC(){ int *s=Que,*e=Que; for(int i=0; i<CH; ++i) if(tree[0][i]) { fail[tree[0][i]]=0;//修改失败指针 *e++=tree[0][i]; } while(s!=e) { int p=*s++; for(int i=0; i<CH; ++i) { if(tree[p][i]) { int v=tree[p][i]; *e++=v; fail[v]=tree[fail[p]][i]; word[v]|=word[fail[v]];//Wa了N次 } else { tree[p][i]=tree[fail[p]][i]; } } }}char s[13];void Input(int m){ fail[0]=0; memset(tree[0],0,sizeof(tree[0]));//初始化根节点，开始都指向根节点 //从1开始，0有特殊使用 cnt=1;//已经有根节点,节点是从1<=i<cnt for(int i=0; i<CH; i++) num[i+'a']=i; for(int i=1; i<=m; i++) { scanf("%s",s); Ins(s,1); }}void init()//建立矩阵的关系{ memset(mat,0,sizeof(mat)); memset(mm,0,sizeof(mm)); memset(matcal,0,sizeof(matcal)); matnum=0;//矩阵从0<=i<matnum for(int i=0; i<cnt; ++i) if(word[i]==0) matcal[i]=matnum++; for(int i=0; i<cnt; ++i) if(word[i]==0) { for(int j=0; j<CH; ++j) { if(word[tree[i][j]]==0)//不是病毒串 { mm[matcal[i]][matcal[tree[i][j]]]++; } } }}void add(ull a[][NODE], ull b[][NODE]){ memset(tem,0,sizeof(tem)); for(int i = 0; i < matnum; ++i) for(int j = 0; j < matnum; ++j) tem[i][j] = a[i][j] + b[i][j];}void Mul(ull a[NODE][NODE],ull b[NODE][NODE]){ memset(tem,0,sizeof(tem)); for(int i=0; i<matnum; ++i) for(int j=0; j<matnum; ++j) for(int k=0; k<matnum; ++k) tem[i][j]=tem[i][j]+a[i][k]*b[k][j];}void solve(ull r[NODE][NODE],int k)//r=mm^k{ ull mt[NODE][NODE]; memcpy(mt,mm,sizeof(mm)); memset(r,0,sizeof(ull)*NODE*NODE); for(int i=0; i<matnum; ++i) r[i][i]=1; while(k) { if(k&1) { Mul(mt,r); memcpy(r,tem,sizeof(tem)); } k>>=1; Mul(mt,mt); memcpy(mt,tem,sizeof(tem)); }}void work(ull dst[][NODE],int k)//dis=a^1+a^2+...+a^k{ if(k==1) { memcpy(dst,mm,sizeof(mm)); return; } ull res[NODE][NODE],x[NODE][NODE],y[NODE][NODE]; solve(x,k>>1); work(y,k>>1); for(int i = 0; i < matnum; ++i) x[i][i]++;//加上前面那一部分 Mul(x,y); memcpy(res,tem,sizeof(tem)); if(k&1) { solve(x,k);//缺少了一个a^k add(res,x);//res=res+x memcpy(res,tem,sizeof(tem)); } memcpy(dst,res,sizeof(res));}ull inc(int k){ ull r=1,t=CH; while(k) { if(k&1) r=r*t; k>>=1; t=t*t; } return r;}ull cal(int n){ if(n==1) return CH; ull x=inc(n>>1)+1;//加上前面那些 ull y=cal(n>>1); ull res=x*y; if(n&1) { res+=inc(n); } return res;}int main(){ int m,n; while(scanf("%d%d",&m,&n)!=EOF) { Input(m); AC(); init(); work(mat,n); ull res=cal(n); for(int i = 0; i < matnum; ++i) res -= mat[0][i]; printf("%I64u/n",res); } return 0;}