BZOJ2580:[USACO]Video Game(AC自动机,DP)

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

HINT

The optimal sequence of buttons in this case is ABACBCB, which gives 4 points--1 from ABA, 1 from ABACB, and 2 from CB.

Solution

用$End[i]$表示$i$结点及其所有的后缀（也就是其$fail$树上所有祖先）的字符串个数。

设$f[i][j]$表示匹配到$i$点，长度为$j$，转移比较显然。

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #define N (1009)
 using namespace std;

 int n,k,cnt,ans,f[N][N];
 int Son[N][],End[N],Fail[N];
 char s[N];
 queue<int>q;

 void Insert(char s[])
 {
     int now=,len=strlen(s);
     for (int i=; i<len; ++i)
     {
         int x=s[i]-'A';
         if (!Son[now][x]) Son[now][x]=++cnt;
         now=Son[now][x];
     }
     ++End[now];
 }

 void Build_Fail()
 {
     for (int i=; i<; ++i)
         if (Son[][i]) q.push(Son[][i]);
     while (!q.empty())
     {
         int now=q.front(); q.pop();
         End[now]+=End[Fail[now]];
         for (int i=; i<; ++i)
         {
             if (!Son[now][i])
             {
                 Son[now][i]=Son[Fail[now]][i];
                 continue;
             }
             Fail[Son[now][i]]=Son[Fail[now]][i];
             q.push(Son[now][i]);
         }
     }
 }

 int main()
 {
     scanf("%d%d",&n,&k);
     for (int i=; i<=n; ++i)
         scanf("%s",s), Insert(s);
     Build_Fail();
     memset(f,-0x7f,sizeof(f));
     f[][]=;
     for (int i=; i<=k; ++i)
         for (int j=; j<=cnt; ++j)
             for (int k=; k<; ++k)
                 f[Son[j][k]][i+]=max(f[Son[j][k]][i+],f[j][i]+End[Son[j][k]]);
     for (int i=; i<=cnt; ++i) ans=max(ans,f[i][k]);
     printf("%d\n",ans);
 }