[POJ1625]Censored!(AC自动机+DP+高精度)
Censored!
Time Limit: 5000MS Memory Limit: 10000K Total Submissions: 10824 Accepted: 2966 Description
The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.
Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.Input
The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10).
The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32).
The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.Output
Output the only integer number -- the number of different sentences freelanders can safely use.Sample Input
2 3 1 ab bbSample Output
5Source
Northeastern Europe 2001, Northern Subregion
同HDU2222，只是需要高精度
感觉可能有个问题，就是这题题目没有规定屏蔽词包含了所有字母，但是好像对解题没有什么影响。
代码用时：1.5h 高精度输出写炸了
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 #define rep(i,l,r) for (int i=l; i<=r; i++)
5 using namespace std;
6
7 const int N=110;
8 int n,m,p,nd,fail[N],b[N],q[N],c[N][N];
9 char s[N],S[N];
10
11 int get(char ch){ rep(i,1,n) if (S[i]==ch) return i; return -1; }
12
13 void ins(char S[]){
14 int x=0,len=strlen(s+1);
15 rep(i,1,len){
16 int k=get(s[i]);
17 if (!c[x][k]) ++nd,c[x][k]=nd;
18 x=c[x][k];
19 }
20 b[x]=1;
21 }
22
23 void getfail(){
24 int st=0,ed=0;
25 rep(i,1,n) if (c[0][i]) q[++ed]=c[0][i];
26 while (st!=ed){
27 int x=q[++st];
28 rep(i,1,n)
29 if (!c[x][i]) c[x][i]=c[fail[x]][i];
30 else q[++ed]=c[x][i],fail[c[x][i]]=c[fail[x]][i];
31 b[x]|=b[fail[x]];
32 }
33 }
34
35 struct D{
36 int v[N],len;
37 D(int x=0){
38 memset(v,0,sizeof(v));
39 for (len=0; x>0; x/=10) v[len++]=x%10;
40 len--;
41 }
42 D operator +(const D &a){
43 D ans;
44 ans.len=max(len,a.len);
45 for (int i=0; i<=ans.len; i++){
46 ans.v[i]+=v[i]+a.v[i]; ans.v[i+1]+=ans.v[i]/10; ans.v[i]%=10;
47 }
48 while (ans.v[ans.len+1]) ans.len++;
49 return ans;
50 }
51 void print(){
52 if (len==-1) { printf("%d\n",0); return; }
53 for (int i=len; ~i; i--) printf("%d",v[i]);
54 printf("\n");
55 }
56 }dp[52][N];
57
58 int check(int k,int j){ int ans=0; rep(i,1,n) if (c[k][i]==j) ans++; return ans; }
59
60 int main(){
61 freopen("poj1625.in","r",stdin);
62 freopen("poj1625.out","w",stdout);
63 scanf("%d%d%d",&n,&m,&p); scanf("%s",S+1);
64 rep(i,1,p) scanf("%s",s+1),ins(s);
65 getfail(); dp[0][0]=D(1);
66 rep(i,1,m) rep(j,0,nd) if (!b[j])
67 rep(k,0,nd) if (!b[k])
68 for (int ti=check(k,j); ti--; ) dp[i][j]=dp[i][j]+dp[i-1][k];
69 D ans; rep(i,0,nd) ans=ans+dp[m][i];
70 ans.print();
71 return 0;
72 }
