poj1625-AC自动机-dp-大数

题目链接：http://poj.org/problem?id=1625

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

题意
给出n个字符,p个违规串，问有多少长度为m的串不包含违规串。

思路
AC自动机加dp,同时转移时数据过大要大数。dp[i][j]记录长度为i时,第i位在AC自动机上编号为j的点时的方案数。转移方程dp[i+1][AC[j][k]]+=dp[i][j]。最后取dp[m][0到AC自动机编号上限]的和。

---

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125

#include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; const int mod = 1e8; struct node { int len; int a[15]; void operator +=(const node &x) { int p = 0; for (int i = 0; i <= len || i <= x.len; i++) { int c = p + a[i] + x.a[i]; a[i] = c % mod; p = c / mod; } len = max(len, x.len); if (p) a[++len] = p; } void print() { printf("%d", a[len]); for (int i = len - 1; i >= 0; i--) printf("%08d", a[i]); cout << endl; } }; int n, m, p; struct Trie { int AC[105][55]; int fail[105]; int numend[105]; node dp[55][105]; int mp[256]; int len; int root; int newcode() { memset(AC[len], 0, sizeof(AC[len])); numend[len++] = 0; return len - 1; } void init() { memset(mp, 0, sizeof(mp)); memset(dp, 0, sizeof(dp)); dp[0][0].a[0] = 1; len = 0; root = newcode(); } void idx(string s) { int len = s.length(); for (int i = 0; i < len; i++) mp[s[i]] = i; } void build(string s) { int len = s.length(); int now = root; for (int i = 0; i < len; i++) { if (!AC[now][mp[s[i]]]) { AC[now][mp[s[i]]] = newcode(); } now = AC[now][mp[s[i]]]; } numend[now] = 1; } void get_fail() { queue<int>q; for (int d = 0; d < n; d++) if (AC[root][d]) { fail[AC[root][d]] = root; q.push(AC[root][d]); } else AC[root][d] = root; while (!q.empty()) { int now = q.front(); q.pop(); numend[now] |= numend[fail[now]]; for (int d = 0; d < n; d++) { if (!AC[now][d]) { AC[now][d] = AC[fail[now]][d]; continue; } fail[AC[now][d]] = AC[fail[now]][d]; q.push(AC[now][d]); } } } void slove() { for (int i = 0; i < m; i++) for (int j = 0; j < len; j++) if (!numend[j]) for (int k = 0; k < n; k++) { if (!numend[AC[j][k]]) dp[i + 1][AC[j][k]]+= dp[i][j]; } node ans; memset(ans.a, 0, sizeof(ans.a)); ans.len = 0; for (int j = 0; j < len; j++) ans += dp[m][j]; ans.print(); } }ac; string s; int main() { while (scanf("%d%d%d", &n, &m, &p) != EOF) { ac.init(); cin >> s; ac.idx(s); for (int i = 0; i < p; i++) { cin >> s; ac.build(s); } ac.get_fail(); ac.slove(); } }

赏

谢谢你请我吃糖果