poj 1625 Censored!
2016-03-22 22:05
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Description
The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.
Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.
Input
The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10).
The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32).
The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.
Output
Output the only integer number -- the number of different sentences freelanders can safely use.
Sample Input
Sample Output
View Code
The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.
Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.
Input
The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10).
The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32).
The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.
Output
Output the only integer number -- the number of different sentences freelanders can safely use.
Sample Input
2 3 1 ab bb
Sample Output
5 [b]换姿势记路径 [/b]
#include<queue> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int bi=1e4,MN=40; char c[1000]; struct big{ int a[MN]; inline big(){ memset(a,0,sizeof(a)); a[0]=1; } inline void read(){ register int i,j; scanf("%s",c); a[0]=(strlen(c)+3)/4; for (i=0;i<strlen(c);i++) j=(strlen(c)-i+3)/4,a[j]=a[j]*10+c[i]-48; } inline void pr(){ register int i; printf("%d",a[a[0]]); for (i=a[0]-1;i;i--) printf("%04d",a[i]); } inline big operator =(int x){ if (x==0){ memset(a,0,sizeof(a)); a[0]=1; } a[0]=0; while (x){ a[0]++; a[a[0]]=x%bi; x/=bi; } if (!a[0]) a[0]=1; } inline big(int x){ *this=x; } inline void gl(){ while(!a[a[0]]&&a[0]>1) a[0]--; } inline big operator =(big x){ register int i; a[0]=x.a[0]; for (i=1;i<=a[0];i++) a[i]=x.a[i]; } inline bool operator >(big y){ if (a[0]!=y.a[0]) return a[0]>y.a[0]; for (register int i=a[0];i;i--){ if (a[i]!=y.a[i]) return a[i]>y.a[i]; } return 0; } inline bool operator >=(const big y){ if (a[0]!=y.a[0]) return a[0]>y.a[0]; for (register int i=a[0];i;i--){ if (a[i]!=y.a[i]) return a[i]>y.a[i]; } return 1; } inline bool operator <(big y){ if (a[0]!=y.a[0]) return a[0]<y.a[0]; for (register int i=a[0];i;i--){ if (a[i]!=y.a[i]) return a[i]<y.a[i]; } return 0; } inline bool operator <=(big y){ if (a[0]!=y.a[0]) return a[0]<y.a[0]; for (register int i=a[0];i;i--){ if (a[i]!=y.a[i]) return a[i]<y.a[i]; } return 1; } inline bool operator ==(big y){ if (a[0]!=y.a[0]) return 0; for (register int i=a[0];i;i--){ if (a[i]!=y.a[i]) return 0; } return 1; } inline bool operator !=(big y){ return !(*this==y); } inline bool operator ==(int y){ big x=y; return *this==x; } inline bool operator !=(int y){ return !(*this==y); } inline void swap(big &a,big &b){ big x=a;a=b;b=x; } inline big operator +(big x){ big r; if (a[0]<x.a[0]) r.a[0]=x.a[0];else r.a[0]=a[0]; for (register int i=1;i<=r.a[0];i++) r.a[i]=a[i]+x.a[i]; for (register int i=1;i<=r.a[0];i++) if (r.a[i]>=bi){ r.a[i]-=bi;r.a[i+1]++; if (i==r.a[0]) r.a[0]++; } return r; } inline big operator -(big x){ if (*this<x) swap(*this,x); register int i; big r; if (a[0]<x.a[0]) r.a[0]=x.a[0];else r.a[0]=a[0]; for (i=1;i<=r.a[0];i++) r.a[i]=a[i]-x.a[i]; for (i=1;i<=r.a[0];i++) if (r.a[i]<0){ r.a[i+1]--;r.a[i]+=bi; } r.gl(); return r; } inline big operator *(big y){ register int i,j; big r;r.a[0]=a[0]+y.a[0]-1; for (i=1;i<=a[0];i++) for (j=1;j<=y.a[0];j++) r.a[i+j-1]+=a[i]*y.a[j]; for (i=0;i<=r.a[0];i++) if (r.a[i]>=bi){ r.a[i+1]+=r.a[i]/bi; r.a[i]%=bi; if (i==r.a[0]) r.a[0]++; } return r; } inline big half(){ register int i,j; for (i=a[0];i>1;i--) a[i-1]+=(a[i]%2)*bi,a[i]/=2; a[1]/=2; gl(); return *this; } inline big operator /(big y){ register int i,j; big r,l,mid,rq=*this; r.a[0]=rq.a[0]+1;r.a[r.a[0]]=1; while(r>l){ mid=(l+r+big(1)).half(); if (mid*y<=rq) l=mid;else r=mid-big(1); } return l; } inline big operator %(big y){ register int i,j; big rq=*this; return rq-(rq/y*y); } inline big operator ^(int y){ big ans=1; big rq=*this; while(y){ if (y&1) ans=ans*rq; y>>=1; rq=rq*rq; } return ans; } inline big operator ^(big y){ big ans=1; big rq=*this; while(y!=0){ if (y.a[1]&1) ans=ans*rq; y=y/2; rq=rq*rq; } return ans; } }; const int LO=51; int nnuu[300]; inline int f(char u){ return nnuu[u]; } struct tree{ int f; bool w; int t[LO]; int v[LO]; }t[101]; int tt,n,m,p,num=0; char s[1000]; bool ma[105],us[105]; queue <int> q; big dp[51][101]; inline bool dfs(int x){ if (x==0) return 1; if (t[x].w) return 0; if (us[x]) return ma[x]; us[x]=1; return ma[x]=dfs(t[x].f); } inline void in(){ int p=0,l,m=strlen(s); for (register int i=0;i<m;i++){ l=f(s[i]); if (!t[p].t[l]) t[p].t[l]=++num; p=t[p].t[l]; } t[p].w=1; } inline void mafa(){ register int i;int k,p; q.push(0);t[0].f=0; while(!q.empty()){ k=q.front();q.pop(); for (i=0;i<n;i++) if (t[k].t[i]){ p=t[k].f; while((!t[p].t[i])&&p) p=t[p].f; t[t[k].t[i]].f=(k==p)?0:t[p].t[i]; q.push(t[k].t[i]); } } } int main(){ register int i,j,k;int u;big ans; while(~scanf("%d%d%d",&n,&m,&p)){ num=0;u=0;ans=0; scanf("%s",s); for (i=0;i<n;i++) nnuu[s[i]]=i; for (i=1;i<=p;i++){ scanf("%s",s); in(); } mafa(); for (i=0;i<=num;i++) ma[i]=dfs(i); dp[0][0]=1; for (i=0;i<=num;i++) if (ma[i]) for (j=0;j<n;j++){ if (!t[i].t[j]){ u=t[i].f; while(!t[u].t[j]&&u)u=t[u].f; u=t[u].t[j]; }else u=t[i].t[j]; t[i].v[j]=u; } for (i=0;i<m;i++) for (j=0;j<=num;j++) if (ma[j]) if (dp[i][j]!=0) for (k=0;k<n;k++) if (ma[t[j].v[k]]) dp[i+1][t[j].v[k]]=dp[i][j]+dp[i+1][t[j].v[k]]; for (i=0;i<=num;i++) if (ma[i]) ans=ans+dp[m][i]; ans.pr();printf("\n"); for (i=0;i<=m;i++) for (j=0;j<=num;j++) dp[i][j]=0; for (i=0;i<=num;i++) for (j=0;j<n;j++) t[i].t[j]=0; for (i=0;i<=num;i++) t[i].w=t[i].f=0; } }
View Code
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