[洛谷3041]视频游戏的连击Video Game Combos

题目描述

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?

贝西在玩一款游戏，该游戏只有三个技能键 “A”“B”“C”可用，但这些键可用形成N种(1 <= N<= 20)特定的组合技。第i个组合技用一个长度为1到15的字符串S_i表示。

当贝西输入的一个字符序列和一个组合技匹配的时候，他将获得1分。特殊的，他输入的一个字符序列有可能同时和若干个组合技匹配，比如N=3时，3种组合技分别为"ABA", "CB", 和"ABACB",若贝西输入"ABACB"，他将获得3分。

若贝西输入恰好K (1 <= K <= 1,000)个字符，他最多能获得多少分？

输入输出格式

输入格式：

* 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.

输出格式：

* Line 1: A single integer, the maximum number of points Bessie can obtain.

输入输出样例

输入样例#1：
复制

3 7
ABA
CB
ABACB

输出样例#1： 复制

4

说明

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

很显然的AC自动机+dp，不过dp那块还是调了半天。。

代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 using namespace std;
 int n,cnt,ans,k;
 int fail[];
 int end[];
 int ch[][];
 int dp[][];
 bool vis[][];
 char s[];
 void build(string s)
 {
     int len=s.length();
     int now=;
     for(int i=;i<len;i++)
     {
         if(!ch[now][s[i]-'A']) ch[now][s[i]-'A']=++cnt;
         now=ch[now][s[i]-'A'];
     }
     end[now]+=;
 }
 void build_fail()
 {
     queue<int>q;
     for(int i=;i<;i++)
         if(ch[][i])
             q.push(ch[][i]);
     while(!q.empty())
     {
         int u=q.front(); q.pop();
         for(int i=;i<;i++)
         {
             if(ch[u][i])
             {
                 fail[ch[u][i]]=ch[fail[u]][i];
                 q.push(ch[u][i]);
             }
             else ch[u][i]=ch[fail[u]][i];
         }
     }
 }
 int get(int now,int val)
 {
     while(now) val+=end[now],now=fail[now];
     return val;
 }
 int main()
 {
     scanf("%d%d",&n,&k);
     for(int i=;i<=n;i++)
     {
         scanf("%s",s);
         build(s);
     }
     build_fail();
     vis[][]=;
     for(int i=;i<k;i++)
         for(int j=;j<=cnt;j++)
         {
             if(!vis[i][j]) continue;
             for(int l=;l<;l++)
             {
                 int now=ch[j][l];
                 dp[i+][now]=max(dp[i+][now],get(now,dp[i][j]));
                 vis[i+][now]=true;
             }
         }
     for(int i=;i<=cnt;i++) ans=max(ans,dp[k][i]);
     printf("%d",ans);
     return ;
 }