给出n个模板串(n<6)求出长度为不超过l(l<2^31)的单词至少包含n个字串中的一个的种类数,对2^64取模。
首先有多个模板串,考虑Aho-Corasick,然后l数据范围提示要用log级别的算法,Trie中最常见的就是矩阵,那么接着分析,问出不超过l至少包含1个,那么我们把问题简化,我们会求出长度为l的不包含任意一个情况吧,不会的同学,传送过去,就用26^l-A^l,就得到长度为l包含至少一个的个数,那么就可以得到总体的算法,26^1+26^2+26^3+26^4+......26^l-(A^1+A^2+A^3+A^4+......A^l),矩阵的幂求和可以这样:
|A,1|^(n+1) = |A^(n+1),1+A^1+A^2+A^3+A^4+......A^l|
|0,1| |0 , 1|
26^1+26^2+26^3+26^4+......26^l = 26*(1-26^l)/(1-26)
#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#define LL unsigned long long
using namespace std;
const int N = 40;
const int alp = 26;
struct node{
int id;
bool flag;
node *ch[alp],*fail;
void init(){
fail = NULL;
for(int i = 0;i < alp;++i)ch[i] = NULL;
}
}trie[N];
char s[250];
int m,ncnt;
LL l;
struct Matrix{
LL map[2*N][2*N];
void clear(){
memset(map,0,sizeof(map));
}
}c;
node *Newnode(){
node *p = &trie[ncnt];
p->init();
p->id = ncnt++;
return p;
}
void insert(node *root,char *s){
node *p = root;
while(*s != ‘\0‘){
if(!p->ch[*s-‘a‘])p->ch[*s-‘a‘] = Newnode();
p = p->ch[*s-‘a‘];
++s;
}
p->flag = true;
}
void _build(node *root){
memset(c.map,0,sizeof(c.map));
queue <node *> q;
q.push(root);
while(!q.empty()){
node *p = q.front();q.pop();
for(int i = 0;i < alp;++i){
if(p->ch[i]){
node *next = p->fail;
while(next && !next->ch[i])next = next->fail;
p->ch[i]->fail = next ? next->ch[i]:root;
if(p->ch[i]->fail->flag)p->ch[i]->flag = true;
q.push(p->ch[i]);
}
else p->ch[i] = (p==root) ? root:p->fail->ch[i];
if(!p->ch[i]->flag)++c.map[p->id][p->ch[i]->id];
}
}
}
Matrix Mul(Matrix x,Matrix y){
Matrix res;
for(int i = 0;i < 2*ncnt;++i){
for(int j = 0;j < 2*ncnt;++j){
res.map[i][j] = 0;
for(int k = 0;k < 2*ncnt;++k){
res.map[i][j] = (res.map[i][j]+y.map[k][j]*x.map[i][k]);
}
}
}
return res;
}
Matrix Pow(Matrix x,LL n){
Matrix res;
res.clear();
for(int i = 0;i < 2*ncnt;++i)res.map[i][i] = 1;
while(n){
if(n&1)res = Mul(res,x);
x = Mul(x,x);
n >>=1;
}
return res;
}
Matrix Sum(Matrix x,LL l){
Matrix ret;
for(int i = 0;i < ncnt;++i){
for(int j = 0;j < ncnt;++j){
ret.map[i][j] = c.map[i][j];
}
}
for(int i = 0;i < ncnt;++i){
for(int j = ncnt;j < 2*ncnt;++j){
ret.map[i][j] = ( i == (j-ncnt));
}
}
for(int i = ncnt;i < 2*ncnt;++i){
for(int j = 0;j < ncnt;++j){
ret.map[i][j] = 0;
}
}
for(int i = ncnt;i < 2*ncnt;++i){
for(int j = ncnt;j < 2*ncnt;++j){
ret.map[i][j] = ( (i-ncnt) == (j-ncnt));
}
}
ret = Pow(ret,l+1);//l+1可能要爆Int
return ret;
}
LL fastp(int x,int n){
LL res = 1;
LL f = (LL)x;
while(n){
if(n&1)res = res*f;
f = f*f;
n >>= 1;
}
return res;
}
void _pre(){
for(int i = 0;i < ncnt;++i){
if(trie[i].flag){
for(int k = 0;k < ncnt;++k)c.map[i][k] = 0;
for(int k = 0;k < ncnt;++k)c.map[k][i] = 0;
}
}
}
int main(){
while(scanf("%d%I64u",&m,&l) != EOF){
ncnt = 0;
memset(trie,0,sizeof(trie));
node *root = Newnode();
for(int i = 0;i < m;++i){
scanf("%s",s);
insert(root,s);
}
_build(root);
LL ans = 0;
_pre();
Matrix res = Sum(c,l);
for(int i = ncnt;i < 2*ncnt;++i){
ans += res.map[0][i];
}
ans -= 1;
ans = (LL)26*(fastp(26,l)-1)*10330176681277348905LL-ans;
//10330176681277348905LL是(x/25)%2^64的x的逆元
cout<<ans<<endl;
}
return 0;
}
hdu 2243 (Aho-Corasick & 矩阵优化幂求和) - xgtao -
原文:http://www.cnblogs.com/xgtao984/p/5701648.html