POJ1625-Censored!

Censored!

Time Limit: 5000MS Memory Limit: 10000KTotal Submissions: 10565 Accepted: 2891

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 1abbb

Sample Output

5

Source

Northeastern Europe 2001, Northern Subregion

题意：给你m个字符和p个字符串，问用这m个字符去构造一个长度为n的且不包含这p个字符串中任意一个的字符串能有几种

解题思路：ac自动机+dp，这个答案会很大，要用大数，所以矩阵快速幂不好弄，用dp做，dp[i][j]表示长度为i,在节点j满足条件的方案数

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cctype>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;struct BigInt{const static int mod = 10000;const static int DLEN = 4;int a[600], len;BigInt(){memset(a, 0, sizeof(a));len = 1;}BigInt(int v){memset(a, 0, sizeof(a));len = 0;do{a[len++] = v%mod;v /= mod;} while (v);}BigInt(const char s[]){memset(a, 0, sizeof(a));int L = strlen(s);len = L / DLEN;if (L%DLEN)len++;int index = 0;for (int i = L - 1; i >= 0; i -= DLEN){int t = 0;int k = i - DLEN + 1;if (k < 0)k = 0;for (int j = k; j <= i; j++)t = t * 10 + s[j] - '0';a[index++] = t;}}BigInt operator +(const BigInt &b)const{BigInt res;res.len = max(len, b.len);for (int i = 0; i <= res.len; i++)res.a[i] = 0;for (int i = 0; i < res.len; i++){res.a[i] += ((i < len) ? a[i] : 0) + ((i < b.len) ? b.a[i] : 0);res.a[i + 1] += res.a[i] / mod;res.a[i] %= mod;}if (res.a[res.len] > 0)res.len++;return res;}BigInt operator *(const BigInt &b)const{BigInt res;for (int i = 0; i < len; i++){int up = 0;for (int j = 0; 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 != 0)res.a[i + b.len] = up;}res.len = len + b.len;while (res.a[res.len - 1] == 0 && res.len > 1)res.len--;return res;}void output(){printf("%d", a[len - 1]);for (int i = len - 2; i >= 0; i--)printf("%04d", a[i]);printf("\n");}}dp[2][110];struct Matrix{int v[110][110];Matrix(){memset(v, 0, sizeof v);}}x;int n, m, p;char ch[100];map<char, int>mp;struct Trie{int next[110][60], fail[110], flag[110];int root, tot;int newnode(){for (int i = 0; i < n; i++) next[tot][i] = -1;flag[tot++] = 0;return tot - 1;}void init(){tot = 0;root = newnode();}void insert(char ch[]){int k = root;for (int i = 0; ch[i]; i++){if (next[k][mp[ch[i]]] == -1) next[k][mp[ch[i]]] = newnode();k = next[k][mp[ch[i]]];}flag[k] = 1;}void build(){queue<int>q;fail[root] = root;for (int i = 0; i < n; i++){if (next[root][i] == -1) next[root][i] = root;else{fail[next[root][i]] = root;q.push(next[root][i]);}}while (!q.empty()){int pre = q.front();q.pop();if (flag[fail[pre]]) flag[pre] = 1;for (int i = 0; i < n; i++){if (next[pre][i] == -1) next[pre][i] = next[fail[pre]][i];else{fail[next[pre][i]] = next[fail[pre]][i];q.push(next[pre][i]);}}}}void solve(){for (int i = 0; i < tot; i++)for (int j = 0; j < n; j++)if (!flag[next[i][j]]) x.v[i][next[i][j]]++;int now = 0;for (int i = 0; i < tot; i++) dp[now][i] = 0;dp[now][0] = 1;for (int i = 1; i <= m; i++){for (int j = 0; j < tot; j++) dp[now ^ 1][j] = 0;for (int j = 0; j < tot; j++)for (int k = 0; k < tot; k++)if (x.v[j][k]) dp[now ^ 1][k] = dp[now ^ 1][k] + dp[now][j] * x.v[j][k];now ^= 1;}BigInt ans = 0;for (int i = 0; i < tot; i++) ans = ans + dp[now][i];ans.output();}void debug(){for (int i = 0; i < tot; i++){printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], flag[i]);for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);printf("]\n");}}}ac;int main(){while (~scanf("%d %d %d", &n, &m, &p)){mp.clear();scanf("%s", ch);for (int i = 0; ch[i]; i++) mp[ch[i]] = i;ac.init();for (int i = 1; i <= p; i++){scanf("%s", &ch);ac.insert(ch);}ac.build();ac.solve();}return 0;}