省队集训day6 B

一道AC自动机题····

一定要把一个节点没有的儿子接到它fai的儿子，否则会卡到n^2的·······

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<ctime>
 #define maxn 1048580
 #define maxl 10005
 using namespace std;
 typedef long long int64;
 char ch,s[maxl];
 int fuckwmj=;
 int n,l,head,tail,list[maxn];
 int64 len;
 struct Map{
     int n,ans,ind[maxn],tot,now[maxn],son[maxn<<],pre[maxn<<],dep[maxn];
     void init(int idx){
         n=idx,ans=tot=;
         memset(ind,,sizeof(ind));
         memset(now,,sizeof(now));
         memset(dep,,sizeof(dep));
     }
     void put(int a,int b){pre[++tot]=now[a],now[a]=tot,son[tot]=b,ind[b]++;}
     void solve(){
         head=,tail=;
         for (int i=;i<=n;i++) if (!ind[i]) list[++tail]=i,dep[i]=;
         while (head<tail){
             int u=list[++head];
             for (int p=now[u],v=son[p];p;p=pre[p],v=son[p]){
                 ind[v]--,dep[v]=max(dep[v],dep[u]+);
                 if (!ind[v]) list[++tail]=v;
             }
         }
         for (int i=;i<=n;i++) if (ind[i]){puts("Yes");return;}
         for (int i=;i<=n;i++) ans=max(ans,dep[i]);
         if (ans>=len) puts("Yes");
         else puts("No");
     }
 }dag;
 struct trie{
     int idx,son[maxn][],fai[maxn];
     bool ok[maxn],bo[maxn];
     void clear(){
         idx=;
         memset(son,,sizeof(son));
         memset(fai,,sizeof(fai));
         memset(ok,,sizeof(ok));
         memset(bo,,sizeof(bo));
     }
     void insert(){
         int p=;
         for (int i=,op=(s[i]=='T');i<=l;p=son[p][op],i++,op=(s[i]=='T'))
             if (!son[p][op]) son[p][op]=++idx;
         ok[p]=;
     }
     void get_fai(){
         head=,tail=,list[]=,fai[]=-,bo[]=;
         while (head<tail){
             int u=list[++head];
             for (int ch=,v=son[u][ch],t;ch<=;v=son[u][++ch])
                 if (v&&!bo[v]){
                     bo[v]=,list[++tail]=v;
                     for (t=fai[u];t>=&&!son[t][ch];t=fai[t]);
                     if (t>=) fai[v]=son[t][ch]; else fai[v]=;
                     if (!son[v][]) son[v][]=son[fai[v]][];
                     if (!son[v][]) son[v][]=son[fai[v]][];
                     ok[v]=ok[v]|ok[u]|ok[fai[v]];
                 }
         }
         dag.init(idx);
         head=,tail=,list[]=,bo[]=;
         memset(bo,,sizeof(bo));
         while (head<tail){
             int u=list[++head];
             for (int ch=,v=son[u][ch];ch<=;v=son[u][++ch])
                 if (v&&!ok[v]){
                     dag.put(u,v);
                     if (!bo[v]) bo[v]=,list[++tail]=v;
                 }
         }
     }
 }T;
 void get(){
     for (ch=getchar();ch!='A'&&ch!='T';ch=getchar());
     for (l=;ch=='A'||ch=='T';s[++l]=ch,ch=getchar());
 }
 int main(){
     while (cin>>n>>len){
         T.clear();
         for (int i=;i<=n;i++) get(),T.insert();
         T.get_fai(),dag.solve();
     }
     return ;
 }