【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);

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 const int INF=-2e8;
 8 struct node
 9 {
10     int next[3];
11     int cnt;
12 }a[505];
13 int f[1005][505];
14 int root=0,num=0,fail[505];
15 char s[1005];
16 void Clear()
17     {
18         a[num].cnt=0;
19         for(int i=0;i<3;i++)a[num].next[i]=0;
20     }
21 void add()
22     {
23         scanf("%s",s);
24         int p=root;
25         for(int i=0,ls=strlen(s);i<ls;i++)
26             {
27                 if(a[p].next[s[i]-'A'])p=a[p].next[s[i]-'A'];
28                 else
29                     {
30                         a[p].next[s[i]-'A']=++num;
31                         Clear();
32                         p=num;
33                     }
34             }
35         a[p].cnt++;
36     }
37 void getfail()
38     {
39         queue<int>q;
40         q.push(root);
41         int u,p,v;
42         while(!q.empty())
43             {
44                 u=q.front();q.pop();
45                 for(int i=0;i<3;i++)
46                     {
47                         if(!a[u].next[i])
48                             {
49                                 if(a[fail[u]].next[i])a[u].next[i]=a[fail[u]].next[i];
50                                 continue;
51                             }
52                         p=fail[u];
53                         while(p)
54                             {
55                                 if(a[p].next[i])break;
56                                 p=fail[p];
57                             }
58                         if(a[p].next[i] && a[p].next[i]!=a[u].next[i])fail[a[u].next[i]]=a[p].next[i];
59                         v=a[u].next[i];
60                         a[v].cnt+=a[fail[v]].cnt;
61                         q.push(a[u].next[i]);
62                     }
63             }
64     }
65 int main()
66     {
67         //freopen("pp.in","r",stdin);
68         int n,k,ans=0;
69         scanf("%d%d",&n,&k);
70         for(int i=1;i<=n;i++)
71             add();
72         getfail();
73         for(int i=0;i<=k;i++)
74             for(int j=0;j<=num;j++)
75                 f[i][j]=INF;
76         f[0][0]=0;
77         for(int i=1;i<=k;i++)
78             {
79                 for(int j=0;j<=num;j++)
80                     {
81                         for(int k=0;k<3;k++)
82                             {
83                                 f[i][a[j].next[k]]=max(f[i][a[j].next[k]],f[i-1][j]+a[a[j].next[k]].cnt);
84                             }
85                     }
86             }
87           for(int i=1;i<=num;i++)if(f[k][i]>ans)ans=f[k][i];
88           printf("%d\n",ans);
89         return 0;
90     }

 

posted @ 2017-07-02 11:51  PIPIBoss  阅读(370)  评论(0)  编辑  收藏  举报