HDU 2825 Wireless Password-AC + DP

Give m words. A new word consisting of m words (two words can overlap) has a length of n and contains no less than k basic words. How many new words are there?

 

M is very small, and binary is used to indicate that a new word contains a basic word.

Using m words to create an AC automatic machine, you can find out the mutual inclusion of all words. The suffix feature of the AC automatic machine (the mismatched edge of each node points to a new node, the string from the new node to the trie root is the suffix of the current node string ).

 

Dynamic Planning:

F (I, j, k): a word with the length of I. k (Binary) is contained at the j node in the trie tree ), total number of solutions.

 

Transfer equation:

Add a letter to the end of the current string.

F (I + 1, u, k | val [u]) + = f (I, j, k)

U is the child node of the j node (or the node pointed by the mismatched edge). k | val [u] indicates the situation where a new letter is added and the basic word is contained.

 

# Include <cstdio> # include <cstring> # include <algorithm> # include <queue> using namespace std; struct AC_Automata {# define N 102 # define M 26 int ch [N] [M], val [N], last [N], f [N], sz; void clear () {sz = 1; memset (ch [0], 0, sizeof (ch [0]);} int idx (char c) {return c-'A';} void insert (char s [], int v) {int u = 0; for (int I = 0; s [I]; I ++) {int c = idx (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] = 1 <v;} 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]; if (! U) {ch [r] [c] = ch [f [r] [c]; val [r] = val [r] | val [f [r]; // If the suffix of a string from the root node to the current node contains basic words (PASS) continue;} q. push (u); f [u] = ch [f [r] [c]; last [u] = val [f [u]? F [u]: last [f [u] ;}}} ac; int c [1030]; /// record each status containing several strings int f [27] [102] [1025]; void init () {memset (c, 0, sizeof (c )); for (int I = 0; I <1024; I ++) {for (int j = 0; j <10; j ++) if (1 <j) & I) c [I] ++ ;}} void solve (int n, int m, int p) {# define Mod 20090717 int u; for (int I = 0; I <= n; I ++) for (int j = 0; j <ac. sz; j ++) for (int k = 0; k <(1 <m); k ++) f [I] [j] [k] = 0; f [0] [0] [0] = 1; for (int I = 0; I <n; I ++) for (int j = 0; j <ac. sz; j ++) {for (int k = 0; k <(1 <m); k ++) {if (f [I] [j] [k] = 0) continue; for (int jj = 0; jj <M; jj ++) {u = ac. ch [j] [jj]; f [I + 1] [u] [ac. val [u] | k] = (f [I + 1] [u] [ac. val [u] | k] + f [I] [j] [k]) % Mod ;}} int ans = 0; for (int I = 0; I <ac. sz; I ++) for (int j = 0; j <(1 <m); j ++) if (c [j]> = p) ans = (ans + f [n] [I] [j]) % Mod; printf ("% d \ n", ans);} int main () {int n, m, k; char s [12]; init (); while (scanf ("% d", & n, & m, & k) = 3) {if (n = 0 & m = 0 & k = 0) break; ac. clear (); for (int I = 0; I <m; I ++) {scanf ("% s", s); ac. insert (s, I);} ac. build (); solve (n, m, k);} return 0 ;}