Description
Bessie is playing a video game! In the game, the three letters 'A', 'B', and 'C' are the only valid buttons. Bessie may press the buttons in any order she likes; however, there are only N distinct combos possible (1 <= N <= 20). Combo i is represented as a string S_i which has a length between 1 and 15 and contains only the letters 'A', 'B', and 'C'. Whenever Bessie presses a combination of letters that matches with a combo, she gets one point for the combo. Combos may overlap with each other or even finish at the same time! For example if N = 3 and the three possible combos are "ABA", "CB", and "ABACB", and Bessie presses "ABACB", she will end with 3 points. Bessie may score points for a single combo more than once. Bessie of course wants to earn points as quickly as possible. If she presses exactly K buttons (1 <= K <= 1,000), what is the maximum number of points she can earn?
给出n个ABC串combo[1..n]和k，现要求生成一个长k的字符串S，问S与word[1..n]的最大匹配数
Input
Line 1: Two space-separated integers: N and K. * Lines 2..N+1: Line i+1 contains only the string S_i, representing combo i.
Output
Line 1: A single integer, the maximum number of points Bessie can obtain.
Sample Input
3 7
ABA
CB
ABACB
Sample Output
4
首先对所有的得分串建立AC自动机，然后考虑dp，设(f[i][j])表示当前长度为(i)，匹配到AC自动机上节点(j)的得分，转移直接枚举(j)之后连的字符即可
然后建fail指针的时候把终止标识符累加起来，这样之后就可以(O(1))询问了
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
int x=0,f=1;char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=3e2,M=1e3;
struct S1{
int trie[N+10][3],fail[N+10],End[N+10],tot,root;
void insert(char *s){
int len=strlen(s),p=root;
for (int i=0;i<len;i++){
if (!trie[p][s[i]-'A']) trie[p][s[i]-'A']=++tot;
p=trie[p][s[i]-'A'];
}
End[p]++;
}
void make_fail(){
static int h[N+10];
int head=1,tail=0;
for (int i=0;i<3;i++) if (trie[root][i]) h[++tail]=trie[root][i];
for (;head<=tail;head++){
int Now=h[head];
End[Now]+=End[fail[Now]];//累计标识符
for (int i=0;i<3;i++){
if (trie[Now][i]){
int son=trie[Now][i];
fail[son]=trie[fail[Now]][i];
h[++tail]=son;
}else trie[Now][i]=trie[fail[Now]][i];
}
}
}
}AC;//Aho-Corasick automation
int f[M+10][N+10];
int main(){
int n=read(),K=read();
for (int i=1;i<=n;i++){
static char s[20];
scanf("%s",s);
AC.insert(s);
}
AC.make_fail();
memset(f,255,sizeof(f));
f[0][0]=0;
for (int i=0;i<K;i++){
for (int j=0;j<=AC.tot;j++){
if (!~f[i][j]) continue;
for (int k=0;k<3;k++){
int tmp=AC.trie[j][k];
f[i+1][tmp]=max(f[i+1][tmp],f[i][j]+AC.End[tmp]);
}
}
}
int Ans=0;
for (int i=0;i<=AC.tot;i++) Ans=max(Ans,f[K][i]);
printf("%d
",Ans);
return 0;
}