给出一个长度为N的小写字母串，现在Mirko有M个若干长度为Li字符串。现在Mirko要用这M个字符串去覆盖给出的那个字符串的。覆盖时，必须保证：
1.Mirko的字符串不能拆开，旋转；
2.Mirko的字符串必须和给出的字符串的某一连续段完全一致才能覆盖，
3.若干次覆盖可以部分重叠
4.Mirko的字符串可以无限使用。
求给出的字符串当中，有多少个字母是无法覆盖的。
小朋友们，作为一名长者，我认为我有必要向你们传授一些人生的经验~；
字符串的一堆函数，慎用慎用；
本人只因没有仔细认真，把strlen(s),strlen(ch)之类的函数写在了循环的里面，造成了无限TLE的惨剧；
图如下：
两个strlen在里面的时候：
#01: Accepted (15ms, 322696KB)
#02: Accepted (15ms, 322696KB)
#03: Accepted (187ms, 322696KB)
#04: Accepted (3234ms, 322696KB)
#05: Accepted (3453ms, 322624KB)
#06: Time Limit Exceeded (?, 322624KB)
#07: Time Limit Exceeded (?, 322624KB)
#08: Time Limit Exceeded (?, 322624KB)
#09: Time Limit Exceeded (?, 322624KB)
#10: Time Limit Exceeded (?, 322624KB)
#11: Time Limit Exceeded (?, 322624KB)
#12: Wrong Answer (5000ms, 322624KB)
#13: Wrong Answer (5000ms, 322624KB)
#14: Time Limit Exceeded (?, 322624KB)
#15: Time Limit Exceeded (?, 322624KB)
#16: Time Limit Exceeded (?, 322624KB)
#17: Wrong Answer (5000ms, 322624KB)
#18: Wrong Answer (5000ms, 322624KB)
#19: Time Limit Exceeded (?, 322624KB)
不知道的会以为我暴力枚举；
一个strlen在循环里的时候：
#01: Accepted (15ms, 345056KB)
#02: Accepted (0ms, 345056KB)
#03: Accepted (0ms, 345056KB)
#04: Accepted (218ms, 345056KB)
#05: Accepted (78ms, 345056KB)
#06: Accepted (93ms, 345056KB)
#07: Accepted (46ms, 345056KB)
#08: Accepted (1656ms, 345056KB)
#09: Time Limit Exceeded (?, 345056KB)
#10: Accepted (93ms, 344984KB)
#11: Accepted (125ms, 344984KB)
#12: Accepted (1562ms, 344984KB)
#13: Time Limit Exceeded (?, 344984KB)
#14: Accepted (171ms, 344984KB)
#15: Accepted (500ms, 344984KB)
#16: Accepted (1640ms, 344984KB)
#17: Wrong Answer (5000ms, 344984KB)
#18: Accepted (250ms, 344984KB)
#19: Accepted (734ms, 344984KB)
还是有三组超了；
没有strlen在里面的时候：
#01: Accepted (0ms, 357712KB)
#02: Accepted (0ms, 357712KB)
#03: Accepted (15ms, 357712KB)
#04: Accepted (78ms, 357712KB)
#05: Accepted (93ms, 357712KB)
#06: Accepted (46ms, 357712KB)
#07: Accepted (46ms, 357712KB)
#08: Accepted (375ms, 357712KB)
#09: Accepted (187ms, 357712KB)
#10: Accepted (93ms, 357712KB)
#11: Accepted (31ms, 357712KB)
#12: Accepted (359ms, 357712KB)
#13: Accepted (328ms, 357712KB)
#14: Accepted (187ms, 357712KB)
#15: Accepted (125ms, 357712KB)
#16: Accepted (359ms, 357712KB)
#17: Accepted (468ms, 357712KB)
#18: Accepted (281ms, 357712KB)
#19: Accepted (140ms, 357712KB)
敲完这道题，我热泪盈眶啊；
同时，我通过这道题敲了好几遍AC自动机模板；
代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<string>
5 #include<cstdlib>
6 #include<ctime>
7 #include<vector>
8 #include<algorithm>
9 #include<queue>
10 #include<map>
11 using namespace std;
12 #define LL long long
13 const int maxn=2500010;
14 int n,m;
15 char s[303000],ch[5050];
16 bool vis[27][27][27][27][27];
17 int linkk[1000000][27],flag[maxn],f[maxn],q[maxn],g[maxn],k[maxn],tail=0,head=1,len=0,next[maxn];
18 void insert(){
19 int now=0,p=strlen(ch);
20 for(int i=0;i<p;i++){
21 if(!linkk[now][ch[i]])linkk[now][ch[i]]=++len,g[len]=ch[i];
22 now=linkk[now][ch[i]];
23 if(i==p-1)flag[now]=p;
24 }
25 }
26 void init(){
27 //printf("first:%d\n",clock());
28 scanf("%d%s%d",&n,s,&m);
29 for(int i=0;i<n;i++)s[i]=s[i]-‘a‘+1;
30 for(int i=4;i<n;i++)vis[s[i-4]][s[i-3]][s[i-2]][s[i-1]][s[i]]=1;
31 //printf("second:%d\n",clock());
32 for(int i=1;i<=m;i++){
33 scanf("%s",ch);
34 bool p=0;
35 int u=strlen(ch);
36 for(int j=0;j<u;j++)ch[j]=ch[j]-‘a‘+1;
37 for(int j=4;j<u;j++)
38 if(!vis[ch[j-4]][ch[j-3]][ch[j-2]][ch[j-1]][ch[j]]){p=1;break;}
39 if(!p)insert();
40 }
41 //printf("init:%d\n",clock());
42 }
43 void bfs(){
44 head=0,tail=0;q[++tail]=0;int x=0,now=0,temp;
45 while(++head<=tail){
46 x=q[head];
47 for(int i=1;i<=26;i++){
48 if(linkk[x][i]){
49 now=linkk[x][i],temp=f[x];
50 if(x){
51 while(temp&&!linkk[temp][i])temp=f[temp];
52 f[now]=linkk[temp][i];
53 if(flag[f[now]])next[now]=f[now];
54 else next[now]=next[f[now]];
55 }
56 q[++tail]=now;
57 }
58 }
59 }
60 }
61 void work(){
62 bfs();
63 //printf("bfs:%d\n",clock());
64 int now=0;
65 for(int i=0;i<n;i++){
66 if(!linkk[now][s[i]]){
67 int temp=f[now];
68 while(!linkk[temp][s[i]]&&temp)temp=f[temp];
69 now=linkk[temp][s[i]];
70 }
71 else now=linkk[now][s[i]];
72 int temp=now;
73 while(temp){
74 if(flag[temp])k[i-flag[temp]+1]=flag[temp];
75 temp=next[temp];
76 }
77 }
78 //printf("work:%d\n",clock());
79 int last=-1,sum=0;
80 for(int i=0;i<n;i++){
81 if(k[i])last=max(last,k[i]+i-1);
82 if(i>last)sum++;
83 }
84 cout<<sum<<endl;
85 //cout<<clock()<<endl;
86 }
87 int main(){
88 freopen("1.in","r",stdin);
89 freopen("1.out","w",stdout);
90 init();
91 work();
92 }