bzoj1030 [JSOI2007]文本生成器

Posted 2020-06-17

题目链接

AC自动机+DP

有点像GT考试，hh[i][j]表示第i为匹配到自动机上j号结点的方案数

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<string>
 7 #include<cmath>
 8 #include<ctime>
 9 #include<queue>
10 #include<stack>
11 #include<map>
12 #include<set>
13 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
14 #define re(i,l,r) for(int i=(l);i<=(r);i++)
15 #define Clear(a,b) memset(a,b,sizeof(a))
16 #define inout(x) printf("%d",(x))
17 #define douin(x) scanf("%lf",&x)
18 #define strin(x) scanf("%s",(x))
19 #define LLin(x) scanf("%lld",&x)
20 #define op operator
21 #define CSC main
22 typedef unsigned long long ULL;
23 typedef const int cint;
24 typedef long long LL;
25 using namespace std;
26 void inin(int &ret)
27 {
28     ret=0;int f=0;char ch=getchar();
29     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=1;ch=getchar();}
30     while(ch>=‘0‘&&ch<=‘9‘)ret*=10,ret+=ch-‘0‘,ch=getchar();
31     ret=f?-ret:ret;
32 }
33 int n,m,ch[6060][26],pre[6060],hh[111][6060],v[6060],ed=1,last[6060];
34 char s[1010];
35 void add(char *s,int x)
36 {
37     int temp=1,len=strlen(s);
38     re(i,0,len-1)
39     {
40         int c=s[i]-‘A‘;
41         if(!ch[temp][c])ch[temp][c]=++ed;
42         temp=ch[temp][c];
43     }
44     v[temp]=x;
45 }
46 void getfail()
47 {
48     queue<int>h;int k;
49     h.push(1);
50     while(!h.empty())
51     {
52         int x=h.front();h.pop();
53         re(i,0,25)
54         {
55             if(!ch[x][i])continue;
56             int vv=ch[x][i];
57             h.push(vv);
58             k=pre[x];
59             while(!ch[k][i])k=pre[k];
60             pre[vv]=ch[k][i];
61             if(v[ch[k][i]])v[ch[x][i]]=v[ch[k][i]];
62             last[vv]=v[pre[vv]]?pre[vv]:last[pre[vv]];
63         }
64     }
65 }
66 cint mod=10007;
67 void dp(int x)
68 {
69     re(i,1,ed)
70     {
71         if(v[i]||!hh[x-1][i])continue;
72         re(j,0,25)
73         {
74             int k=i;
75             while(!ch[k][j])k=pre[k];
76             (hh[x][ch[k][j]]+=hh[x-1][i])%=mod;
77         }
78     }
79 }
80 int main()
81 {
82     inin(n),inin(m);
83     re(i,0,25)ch[0][i]=1;
84     re(i,1,n)
85     {
86         strin(s);
87         add(s,i);
88     }
89     getfail();
90     hh[0][1]=1;
91     re(i,1,m)dp(i);
92     int ans1=0,ans2=1;
93     re(i,1,m)ans2=ans2*26%mod;
94     re(i,1,ed)if(!v[i])(ans1+=hh[m][i])%=mod;
95     printf("%d",(ans2-ans1+mod)%mod);
96      return 0;
97 }