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·······

 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 }