Censored!
| Time Limit: 5000MS | Memory Limit: 10000K | |
| Total Submissions: 6956 | Accepted: 1887 |
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
bb
Sample Output
5
Source
Northeastern Europe 2001, Northern Subregion
很好的题目。。。
注意字符貌似会有超过128的，用map转化下就好了。
然后就是用AC自动机实现状态表。
然后高精度 进行简单的DP；
//============================================================================
// Name : POJ.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================ #include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <map>
using namespace std;
map<char,int>mp;
int N,M,P;
struct Matrix
{
int mat[][];
int n;
Matrix(){}
Matrix(int _n)
{
n=_n;
for(int i = ;i < n;i++)
for(int j = ;j < n;j++)
mat[i][j] = ;
}
};
struct Trie
{
int next[][],fail[];
bool end[];
int L,root;
int newnode()
{
for(int i = ;i < ;i++)
next[L][i] = -;
end[L++] = false;
return L-;
}
void init()
{
L = ;
root = newnode();
}
void insert(char buf[])
{
int len = strlen(buf);
int now = root;
for(int i = ;i < len;i++)
{
if(next[now][mp[buf[i]]] == -)
next[now][mp[buf[i]]] = newnode();
now = next[now][mp[buf[i]]];
}
end[now] = true;
}
void build()
{
queue<int>Q;
fail[root] = root;
for(int i = ;i < ;i++)
if(next[root][i] == -)
next[root][i] = root;
else
{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
while(!Q.empty())
{
int now = Q.front();
Q.pop();
if(end[fail[now]]==true)end[now]=true;
for(int i = ;i < ;i++)
if(next[now][i] == -)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
Matrix getMatrix()
{
Matrix res = Matrix(L);
for(int i = ;i < L;i++)
for(int j = ;j < N;j++)
if(end[next[i][j]]==false)
res.mat[i][next[i][j]]++;
return res;
}
void debug()
{
for(int i = ;i < L;i++)
{
printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
for(int j = ;j < ;j++)
printf("%2d",next[i][j]);
printf("]\n");
}
} }; /*
* 高精度，支持乘法和加法
*/
struct BigInt
{
const static int mod = ;
const static int DLEN = ;
int a[],len;
BigInt()
{
memset(a,,sizeof(a));
len = ;
}
BigInt(int v)
{
memset(a,,sizeof(a));
len = ;
do
{
a[len++] = v%mod;
v /= mod;
}while(v);
}
BigInt(const char s[])
{
memset(a,,sizeof(a));
int L = strlen(s);
len = L/DLEN;
if(L%DLEN)len++;
int index = ;
for(int i = L-;i >= ;i -= DLEN)
{
int t = ;
int k = i - DLEN + ;
if(k < )k = ;
for(int j = k;j <= i;j++)
t = t* + s[j] - '';
a[index++] = t;
}
}
BigInt operator +(const BigInt &b)const
{
BigInt res;
res.len = max(len,b.len);
for(int i = ;i <= res.len;i++)
res.a[i] = ;
for(int i = ;i < res.len;i++)
{
res.a[i] += ((i < len)?a[i]:)+((i < b.len)?b.a[i]:);
res.a[i+] += res.a[i]/mod;
res.a[i] %= mod;
}
if(res.a[res.len] > )res.len++;
return res;
}
BigInt operator *(const BigInt &b)const
{
BigInt res;
for(int i = ; i < len;i++)
{
int up = ;
for(int j = ;j < b.len;j++)
{
int temp = a[i]*b.a[j] + res.a[i+j] + up;
res.a[i+j] = temp%mod;
up = temp/mod;
}
if(up != )
res.a[i + b.len] = up;
}
res.len = len + b.len;
while(res.a[res.len - ] == &&res.len > )res.len--;
return res;
}
void output()
{
printf("%d",a[len-]);
for(int i = len-;i >= ;i--)
printf("%04d",a[i]);
printf("\n");
}
};
char buf[];
BigInt dp[][];
Trie ac;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout); while(scanf("%d%d%d",&N,&M,&P)==)
{
gets(buf);
gets(buf);
mp.clear();
int len = strlen(buf);
for(int i = ;i < len;i++)
mp[buf[i]]=i;
ac.init();
for(int i = ;i < P;i++)
{
gets(buf);
ac.insert(buf);
}
ac.build();
// ac.debug();
Matrix a= ac.getMatrix(); // for(int i = 0;i <a.n;i++)
// {
// for(int j=0;j<a.n;j++)printf("%d ",a.mat[i][j]);
// cout<<endl;
// } int now = ;
dp[now][] = ;
for(int i = ;i < a.n;i++)
dp[now][i] = ;
for(int i = ;i < M;i++)
{
now^=;
for(int j = ;j < a.n;j++)
dp[now][j] = ;
for(int j = ;j < a.n;j++)
for(int k = ;k < a.n;k++)
if(a.mat[j][k] > )
dp[now][k] = dp[now][k]+dp[now^][j]*a.mat[j][k];
// for(int j = 0;j < a.n;j++)
// dp[now][j].output();
}
BigInt ans = ;
for(int i = ;i < a.n;i++)
ans = ans + dp[now][i];
ans.output();
}
return ;
}