給n個字母,構成長度爲m的串,總共有n^m種。給p個字符串,問n^m種字符串中不包含(不是子串)這p個字符串的個數。
將p個不能包含的字符串建立AC自動機,每個結點用val值來標記以當前節點爲後綴的字符串是否包含非法字符串(p個字符串中的任何一個)。
狀態轉移方程:f(i, j) += f(i-1, k)
f(i, j)表示長度爲i的字符串,結尾爲字符j,方程j和k的關係可以從自動機中失配關係直接獲得(j是k的後繼結點)。
最終的結果非常大,因此要用大數,大數模板:http://blog.csdn.net/yang_7_46/article/details/9897563
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef unsigned char uchar;
struct AC_Automata {
#define N 102
int ch[N][55], val[N], last[N], f[N], sz;
void clear() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }
int hash[256], M;
void set_hash(int n, uchar s[]) {
M = n; for (int i=0; i<n; i++) hash[s[i]] = i;
}
void insert(uchar s[], int v) {
int u = 0;
for (int i=0; s[i]; i++) {
int c = hash[s[i]];
if (!ch[u][c]) {
memset(ch[sz], 0, sizeof(ch[sz]));
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
val[u] = v; //標記當前串爲非法的
}
void build() {
queue<int> q;
f[0] = 0;
for (int c=0; c<M; c++) {
int u = ch[0][c];
if (u) { f[u] = last[u] = 0; q.push(u); }
}
while (!q.empty()) {
int r = q.front(); q.pop();
for (int c=0; c<M; c++) {
int u = ch[r][c];
val[r] = val[r] || val[f[r]]; //判斷當前結點是否有非法後綴
if (!u) {
ch[r][c] = ch[f[r]][c];
continue;
}
q.push(u);
f[u] = ch[f[r]][c];
last[u] = val[f[u]] ? f[u] : last[f[u]];
}
}
}
} ac;
struct BigInteger{
int A[25];
enum{MOD = 10000};
BigInteger(){memset(A, 0, sizeof(A)); A[0]=1;}
void set(int x){memset(A, 0, sizeof(A)); A[0]=1; A[1]=x;}
void print(){
printf("%d", A[A[0]]);
for (int i=A[0]-1; i>0; i--){
if (A[i]==0){printf("0000"); continue;}
for (int k=10; k*A[i]<MOD; k*=10) printf("0");
printf("%d", A[i]);
}
printf("\n");
}
int& operator [] (int p) {return A[p];}
const int& operator [] (int p) const {return A[p];}
BigInteger operator + (const BigInteger& B){
BigInteger C;
C[0]=max(A[0], B[0]);
for (int i=1; i<=C[0]; i++)
C[i]+=A[i]+B[i], C[i+1]+=C[i]/MOD, C[i]%=MOD;
if (C[C[0]+1] > 0) C[0]++;
return C;
}
BigInteger operator * (const BigInteger& B){
BigInteger C;
C[0]=A[0]+B[0];
for (int i=1; i<=A[0]; i++)
for (int j=1; j<=B[0]; j++){
C[i+j-1]+=A[i]*B[j], C[i+j]+=C[i+j-1]/MOD, C[i+j-1]%=MOD;
}
if (C[C[0]] == 0) C[0]--;
return C;
}
};
int n, m, p;
uchar s[55];
int main() {
while (scanf("%d %d %d ", &n, &m, &p) == 3) {
ac.clear();
cin >> s; ac.set_hash(n, s);
while (p--) {
cin >> s; ac.insert(s, 1);
}
ac.build();
BigInteger f[51][101];
f[0][0].set(1);
for (int i=1; i<=m; i++)
for (int j=0; j<ac.sz; j++)
for (int k=0; k<n; k++) {
int u = ac.ch[j][k];
if (!ac.val[u]) f[i][u] = f[i][u] + f[i-1][j];
}
BigInteger ans;
for (int i=0; i<ac.sz; i++)
if (!ac.val[i]) ans = ans + f[m][i];
ans.print();
}
return 0;
}