【USACO12JAN】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：

3 7
ABA
CB
ABACB

输出样例#1：

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.

题解：

记cnt[i]为节点i沿着fail一直走下去可以获得的积分,那么

f[i][j]为走了i步到节点j的最大积分 注意初始化...

f[i][a[j].next[k]]=max(f[i][a[j].next[k]],f[i-1][j]+a[a[j].next[k]].cnt);

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #include<queue>
 using namespace std;
 const int INF=-2e8;
 struct node
 {
     int next[];
     int cnt;
 }a[];
 int f[][];
 int root=,num=,fail[];
 char s[];
 void Clear()
     {
         a[num].cnt=;
         for(int i=;i<;i++)a[num].next[i]=;
     }
 void add()
     {
         scanf("%s",s);
         int p=root;
         for(int i=,ls=strlen(s);i<ls;i++)
             {
                 if(a[p].next[s[i]-'A'])p=a[p].next[s[i]-'A'];
                 else
                     {
                         a[p].next[s[i]-'A']=++num;
                         Clear();
                         p=num;
                     }
             }
         a[p].cnt++;
     }
 void getfail()
     {
         queue<int>q;
         q.push(root);
         int u,p,v;
         while(!q.empty())
             {
                 u=q.front();q.pop();
                 for(int i=;i<;i++)
                     {
                         if(!a[u].next[i])
                             {
                                 if(a[fail[u]].next[i])a[u].next[i]=a[fail[u]].next[i];
                                 continue;
                             }
                         p=fail[u];
                         while(p)
                             {
                                 if(a[p].next[i])break;
                                 p=fail[p];
                             }
                         if(a[p].next[i] && a[p].next[i]!=a[u].next[i])fail[a[u].next[i]]=a[p].next[i];
                         v=a[u].next[i];
                         a[v].cnt+=a[fail[v]].cnt;
                         q.push(a[u].next[i]);
                     }
             }
     }
 int main()
     {
         //freopen("pp.in","r",stdin);
         int n,k,ans=;
         scanf("%d%d",&n,&k);
         for(int i=;i<=n;i++)
             add();
         getfail();
         for(int i=;i<=k;i++)
             for(int j=;j<=num;j++)
                 f[i][j]=INF;
         f[][]=;
         for(int i=;i<=k;i++)
             {
                 for(int j=;j<=num;j++)
                     {
                         for(int k=;k<;k++)
                             {
                                 f[i][a[j].next[k]]=max(f[i][a[j].next[k]],f[i-][j]+a[a[j].next[k]].cnt);
                             }
                     }
             }
           for(int i=;i<=num;i++)if(f[k][i]>ans)ans=f[k][i];
           printf("%d\n",ans);
         return ;
     }