題目描述

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?

貝西在玩一款游戲，該游戲只有三個技能鍵 “A”“B”“C”可用，但這些鍵可用形成N種(1 <= N<= 20)特定的組合技。第i個組合技用一個長度為1到15的字符串S_i表示。

當貝西輸入的一個字符序列和一個組合技匹配的時候，他將獲得1分。特殊的，他輸入的一個字符序列有可能同時和若干個組合技匹配，比如N=3時，3種組合技分別為"ABA", "CB", 和"ABACB",若貝西輸入"ABACB"，他將獲得3分。

若貝西輸入恰好K (1 <= K <= 1,000)個字符，他最多能獲得多少分？

輸入輸出格式

輸入格式：

* 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.

輸出格式：

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

輸入輸出樣例

輸入樣例#1： 復制 3 7 ABA CB ABACB 輸出樣例#1： 復制 4

說明

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

?

很顯然的AC自動機+dp，不過dp那塊還是調了半天。。

代碼：

1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 using namespace std; 6 int n,cnt,ans,k; 7 int fail[3010]; 8 int end[3010]; 9 int ch[3010][27]; 10 int dp[3010][1001]; 11 bool vis[3010][1001]; 12 char s[100]; 13 void build(string s) 14 { 15 int len=s.length(); 16 int now=0; 17 for(int i=0;i<len;i++) 18 { 19 if(!ch[now][s[i]-'A']) ch[now][s[i]-'A']=++cnt; 20 now=ch[now][s[i]-'A']; 21 } 22 end[now]+=1; 23 } 24 void build_fail() 25 { 26 queue<int>q; 27 for(int i=0;i<3;i++) 28 if(ch[0][i]) 29 q.push(ch[0][i]); 30 while(!q.empty()) 31 { 32 int u=q.front(); q.pop(); 33 for(int i=0;i<3;i++) 34 { 35 if(ch[u][i]) 36 { 37 fail[ch[u][i]]=ch[fail[u]][i]; 38 q.push(ch[u][i]); 39 } 40 else ch[u][i]=ch[fail[u]][i]; 41 } 42 } 43 } 44 int get(int now,int val) 45 { 46 while(now) val+=end[now],now=fail[now]; 47 return val; 48 } 49 int main() 50 { 51 scanf("%d%d",&n,&k); 52 for(int i=1;i<=n;i++) 53 { 54 scanf("%s",s); 55 build(s); 56 } 57 build_fail(); 58 vis[0][0]=1; 59 for(int i=0;i<k;i++) 60 for(int j=0;j<=cnt;j++) 61 { 62 if(!vis[i][j]) continue; 63 for(int l=0;l<3;l++) 64 { 65 int now=ch[j][l]; 66 dp[i+1][now]=max(dp[i+1][now],get(now,dp[i][j])); 67 vis[i+1][now]=true; 68 } 69 } 70 for(int i=0;i<=cnt;i++) ans=max(ans,dp[k][i]); 71 printf("%d",ans); 72 return 0; 73 }

?

轉載于:https://www.cnblogs.com/Slrslr/p/9496647.html

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