Censored! POJ - 1625 ac自動機+dp+高精度

一、內容

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

二、思路

• 由於方案數很多 故應該使用高精度。 還有注意輸入的字符超出了char的範圍會變成負數。

三、代碼

#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <vector>
using namespace std;
const int N = 55, M = 110;
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, len, tr[M][N], ne[M], fail[M];
BigInteger dp[N][M];
char s[1000];
int mp[1000]; 
void add() {
	int p = 0;
	for (int i = 0; s[i]; i++) {
		int j = mp[(int)s[i] + 200];
		if (!tr[p][j]) tr[p][j] = ++len;
		p = tr[p][j];
	}
	fail[p] = 1;
}
void build() {
	queue<int> q;
	for (int j = 0; j < n; j++) {
		if (tr[0][j]) q.push(tr[0][j]);
	}
	while (!q.empty()) {
		int p = q.front(); q.pop();
		for (int j = 0; j < n; j++) {
			int c = tr[p][j];
			if (!c) tr[p][j] = tr[ne[p]][j];
			else {
				ne[c] = tr[ne[p]][j];
				fail[c] |= fail[ne[c]];
				q.push(c);
			}	
		}
	}
} 
 
int main() {
	scanf("%d%d%d", &n, &m, &P);
	scanf("%s", s);
	for (int i = 0; i < n; i++) {
		mp[(int)s[i] + 200] = i;
	}	
	for (int i = 1; i <= P; i++) {scanf("%s", s); add();}
	build();
	//進行dp求方案數 ]
	dp[0][0].set(1);
	for (int i = 1; i <= m; i++) {
		for (int j = 0; j <= len; j++) {
			for (int k = 0; k < n; k++) {
				int tj = tr[j][k]; //(i - 1, j)-->(i, tj)下一狀態跳到哪兒 
				if (!fail[tj]) dp[i][tj] = dp[i][tj] + dp[i - 1][j];
			}
		} 
	}
	BigInteger ans;
	for (int j = 0; j <= len; j++) ans = ans + dp[m][j];
	ans.print();
	return 0;
}