BZOJ1030: [JSOI2007] Text generator

"Portal: BZOJ1030" Brief test Instructions:

Give n words, ask for the number of strings with at least one word in length n

The following:

Obviously it's hard to find out, so we'll find out the total number of strings, and then find the number of strings that don't contain any one word.

Then subtract is the answer.

It's like a GT exam, but it turns into a bunch of strings, so use AC automata

Set F[I][J] is the number of strings without any one word to the first J point of the dictionary tree when the length is I

Just make sure the state is reasonable and you can inherit the

Reference Code:

#include <cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>using namespacestd;intMod=10007;structtrie{intc[ -],s,fail;} t[11000];intTot,root;voidCleanintx) {memset (t[x].c,-1,sizeof(T[X].C)); T[x].s=0;}Charst[ the];voidbt () {intx=0, Len=strlen (st+1);  for(intI=1; i<=len;i++)    {        inty=st[i]-'A'+1; if(t[x].c[y]==-1) Clean (++tot), t[x].c[y]=tot; X=T[x].c[y]; } T[X].S=1;}intlist[11000];voidBFs () {list[1]=0; intHead=1, tail=1;  while(head<=tail) {        intx=List[head];  for(intI=1; i<= -; i++)        {            intson=T[x].c[i]; if(son==-1)Continue; if(x==0) t[son].fail=0; Else            {                intj=T[x].fail;  while(j!=0&&t[j].c[i]==-1) j=T[j].fail; T[son].fail=max (0, T[j].c[i]); if(t[t[son].fail].s==1) t[son].s=1; } list[++tail]=Son; } head++; }}intf[ the][11000];intMain () {intn,m; scanf ("%d%d",&n,&m); Tot=0; Clean (0);  for(intI=1; i<=n;i++) {scanf ("%s", st+1);    BT ();    } BFS (); Memset (F,0,sizeof(f)); intsum=1;  for(intI=1; i<=m;i++) sum= (sum* -)%Mod; f[0][0]=1;  for(intu=1; u<=m;u++)    {         for(intI=0; i<=tot;i++)        {            if(t[i].s==0&&f[u-1][i]>0)            {                 for(intj=1; j<= -; j + +)                {                    intk=i;  while(k!=0&&t[k].c[j]==-1) k=T[k].fail; intson=T[k].c[j]; if(son==-1) son=0; F[u][son]= (f[u][son]+f[u-1][i])%Mod; }            }        }    }    intans=0;  for(intI=0; i<=tot;i++)if(t[i].s==0) ans= (Ans+f[m][i])%Mod; printf ("%d\n", ((Sum-ans)%mod+mod)%Mod); return 0;}

BZOJ1030: [JSOI2007] Text generator