POJ 1625

来源：互联网发布：seo如何快速k掉网站编辑：程序博客网时间：2024/04/30 19:57

AC自动机 + DP + 高精度

一开始开始犯蠢用矩阵相乘……好不容易搞定爆栈的问题又超时……后来发现还是直接dp的时间复杂度低……

用 F[i][j] 表示 i 长的单词到 j 状态有多少种方法，则有 F[i][j] = sum{ F[i-1][k] } (k->j 可行且j没有被标记)

#include <iostream>#include <algorithm>#include <stdlib.h>#include <stdio.h>using namespace std;#defineFOR(i,n)for(int i=0; i<n; i++)#define CC(A, x)memset(A, x, sizeof(A))const int NN = 200; //最大节点数int CH;  //儿子个数struct Trie{inline int sw(unsigned char *c) {return hash[*c];}void init(){CC(word,0); CC(fail,0); CC(chd, 0); sz = 1;}void read(int N, int P){unsigned char ch;for (int i=0; i<N; i++){cin>>ch; hash[ch]=i;}unsigned char str[16];for (int i=0; i<P; i++){cin>>str; insert(str, 1);}}void insert(unsigned char *s, int v){int p = 0;for (; *s; p=chd[p][sw(s++)])if (!chd[p][sw(s)]) chd[p][sw(s)] = sz++;word[p] += v;}void AC(){int *b=Que, *e=b;FOR(i,CH) if(chd[0][i]) *e++ = chd[0][i];for (int p=*b; b!=e; p=*++b){FOR(i,CH) if(chd[p][i]){fail[*e++ = chd[p][i]] = chd[fail[p]][i];if (word[fail[chd[p][i]]]) word[chd[p][i]] = 1;//传递病毒包含的病毒 }else chd[p][i] = chd[fail[p]][i];}}int hash[256];int chd[NN][50], sz;//节点数组，个数 int word[NN], fail[NN], Que[NN];}trie;struct Number{Number(){CC(A,0); A[0]=1;} void set(int x){CC(A,0); A[0]=1; A[1]=x;}void print(){cout<<A[A[0]];for (int i=A[0]-1; i>0; i--){if (A[i]==0){cout<<"0000"; continue;}for (int k=10; k*A[i]<MOD; k*=10) cout<<'0';cout<<A[i];}cout<<endl;}Number operator + (const Number& B){Number 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;}Number operator * (const Number& B){Number 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& operator [] (int p) {return A[p];}const int& operator [] (int p) const {return A[p];}int A[25];enum{MOD = 10000};};Number F[51][101];int M, P;void DP(){F[0][0].set(1);for (int i=1; i<=M; i++)for (int j=0; j<trie.sz; j++)for (int k=0; k<CH; k++){int id = trie.chd[j][k];if (!trie.word[id]) F[i][id] = F[i][id] + F[i-1][j];}Number ans;for (int i=0; i<trie.sz; i++)if (!trie.word[i]) ans = ans + F[M][i];ans.print();}int main(){trie.init();cin>>CH>>M>>P;trie.read(CH, P);trie.AC();DP();return 0;}