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：

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• 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.

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輸出格式：

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

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輸入輸出樣例

輸入樣例#1：?
3 7
ABA
CB
ABACB 輸出樣例#1：?
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.


最簡單的AC自動機+DP了。。。。再不會的話AC自動機白學了。

#include<bits/stdc++.h> #define ll long long #define maxn 1005 using namespace std; int ch[maxn][3],n,m,k,ans=0; int root=0,tot=0,val[maxn]; int f[maxn],g[maxn][maxn]; char s[maxn];inline int id(char c){return c-'A'; }inline void ins(){int len=strlen(s),now=root;for(int i=0;i<len;i++){int c=id(s[i]);if(!ch[now][c]) ch[now][c]=++tot;now=ch[now][c];}val[now]=1; }inline void get_fail(){queue<int> q;for(int i=0;i<3;i++) if(ch[0][i]){q.push(ch[0][i]);}int r,v,x;while(!q.empty()){x=q.front(),q.pop();for(int i=0;i<3;i++){r=ch[x][i];if(!r){ch[x][i]=ch[f[x]][i];continue;}q.push(r);f[r]=ch[f[x]][i];val[r]+=val[f[r]];}} }inline void dp(){memset(g,-0x3f,sizeof(g));g[0][0]=0;int to;for(int i=0;i<k;i++)for(int j=0;j<=tot;j++)for(int u=0;u<3;u++){to=ch[j][u];g[i+1][to]=max(g[i+1][to],g[i][j]+val[to]);}for(int i=0;i<=tot;i++) ans=max(ans,g[k][i]); }int main(){scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){scanf("%s",s);ins();}get_fail();dp();printf("%d\n",ans);return 0; }

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轉載于:https://www.cnblogs.com/JYYHH/p/8280569.html

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