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·······
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 #include<ctime> 6 #define maxn 1048580 7 #define maxl 10005 8 using namespace std; 9 typedef long long int64; 10 char ch,s[maxl]; 11 int fuckwmj=0; 12 int n,l,head,tail,list[maxn]; 13 int64 len; 14 struct Map{ 15 int n,ans,ind[maxn],tot,now[maxn],son[maxn<<1],pre[maxn<<1],dep[maxn]; 16 void init(int idx){ 17 n=idx,ans=tot=0; 18 memset(ind,0,sizeof(ind)); 19 memset(now,0,sizeof(now)); 20 memset(dep,0,sizeof(dep)); 21 } 22 void put(int a,int b){pre[++tot]=now[a],now[a]=tot,son[tot]=b,ind[b]++;} 23 void solve(){ 24 head=0,tail=0; 25 for (int i=0;i<=n;i++) if (!ind[i]) list[++tail]=i,dep[i]=0; 26 while (head<tail){ 27 int u=list[++head]; 28 for (int p=now[u],v=son[p];p;p=pre[p],v=son[p]){ 29 ind[v]--,dep[v]=max(dep[v],dep[u]+1); 30 if (!ind[v]) list[++tail]=v; 31 } 32 } 33 for (int i=0;i<=n;i++) if (ind[i]){puts("Yes");return;} 34 for (int i=0;i<=n;i++) ans=max(ans,dep[i]); 35 if (ans>=len) puts("Yes"); 36 else puts("No"); 37 } 38 }dag; 39 struct trie{ 40 int idx,son[maxn][2],fai[maxn]; 41 bool ok[maxn],bo[maxn]; 42 void clear(){ 43 idx=0; 44 memset(son,0,sizeof(son)); 45 memset(fai,0,sizeof(fai)); 46 memset(ok,0,sizeof(ok)); 47 memset(bo,0,sizeof(bo)); 48 } 49 void insert(){ 50 int p=0; 51 for (int i=1,op=(s[i]=='T');i<=l;p=son[p][op],i++,op=(s[i]=='T')) 52 if (!son[p][op]) son[p][op]=++idx; 53 ok[p]=1; 54 } 55 void get_fai(){ 56 head=0,tail=1,list[1]=0,fai[0]=-1,bo[0]=1; 57 while (head<tail){ 58 int u=list[++head]; 59 for (int ch=0,v=son[u][ch],t;ch<=1;v=son[u][++ch]) 60 if (v&&!bo[v]){ 61 bo[v]=1,list[++tail]=v; 62 for (t=fai[u];t>=0&&!son[t][ch];t=fai[t]); 63 if (t>=0) fai[v]=son[t][ch]; else fai[v]=0; 64 if (!son[v][0]) son[v][0]=son[fai[v]][0]; 65 if (!son[v][1]) son[v][1]=son[fai[v]][1]; 66 ok[v]=ok[v]|ok[u]|ok[fai[v]]; 67 } 68 } 69 dag.init(idx); 70 head=0,tail=1,list[1]=0,bo[0]=1; 71 memset(bo,0,sizeof(bo)); 72 while (head<tail){ 73 int u=list[++head]; 74 for (int ch=0,v=son[u][ch];ch<=1;v=son[u][++ch]) 75 if (v&&!ok[v]){ 76 dag.put(u,v); 77 if (!bo[v]) bo[v]=1,list[++tail]=v; 78 } 79 } 80 } 81 }T; 82 void get(){ 83 for (ch=getchar();ch!='A'&&ch!='T';ch=getchar()); 84 for (l=0;ch=='A'||ch=='T';s[++l]=ch,ch=getchar()); 85 } 86 int main(){ 87 while (cin>>n>>len){ 88 T.clear(); 89 for (int i=1;i<=n;i++) get(),T.insert(); 90 T.get_fai(),dag.solve(); 91 } 92 return 0; 93 }