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：
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.

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題解：

記cnt[i]為節點i沿著fail一直走下去可以獲得的積分,那么

f[i][j]為走了i步到節點j的最大積分 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 }

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轉載于:https://www.cnblogs.com/Yuzao/p/7105592.html

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