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?
给你个模式串(每个长度≤15,1≤N≤20),串中只含有“ABC”三种字母。求一长度为K(1≤K≤1000)的字符串,使得匹配数最大(重复匹配计多次),输出最大值。
输入
* 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
* Line 1: A single integer, the maximum number of points Bessie can obtain.
样例输入
3 7 ABA CB ABACB
样例输出
4
提示
The optimal sequence of buttons in this case is ABACBCB, which gives 4 points--1 from ABA, 1 from ABACB, and 2 from CB.
dp思想非常明显吧。看到多模式串匹配,AC自动机非常明显吧。结合一下,就做好了!
dp[i][j]表示长度为i时末节点编号为j的最大匹配数。
状态转移方程也比较好想。f[i+1][AC[j].son[t]]=max(f[i+1][AC[j].son[t]],f[i][j]+AC[AC[j].son[t]].val);
ans=max(dp[k][j])。
代码中的AC[i].val表示加入这个节点会增加多少个匹配数。
Code:
#include<bits/stdc++.h> #define N 1005 using namespace std; struct node { int fail,son[5],val; }AC[N]; int n,k,num,ans,Q[N],f[N][N]; char s[N]; void insert(char *s) { int len=strlen(s); int now=0; for(int i=0;i<len;i++) { if(!AC[now].son[s[i]-'A'])AC[now].son[s[i]-'A']=++num; now=AC[now].son[s[i]-'A']; } AC[now].val++; } void getfail() { int head=1,tail=0; AC[0].fail=0; for(int i=0;i<3;i++) { int u=AC[0].son[i]; if(u!=0) { Q[++tail]=u; AC[u].fail=0; } } while(head<=tail) { int now=Q[head]; AC[now].val+=AC[AC[now].fail].val; for(int i=0;i<3;i++) { int u=AC[now].son[i]; if(u!=0) { AC[u].fail=AC[AC[now].fail].son[i]; Q[++tail]=u; }else AC[now].son[i]=AC[AC[now].fail].son[i]; } head++; } } int main() { scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) { scanf("%s",s); insert(s); } getfail(); memset(f,-0x3f,sizeof(f)); f[0][0]=0; ans=0; for(int i=0;i<k;i++) for(int j=0;j<=num;j++) for(int t=0;t<3;t++) f[i+1][AC[j].son[t]]=max(f[i+1][AC[j].son[t]],f[i][j]+AC[AC[j].son[t]].val); for(int i=0;i<=num;i++) ans=max(ans,f[k][i]); printf("%d\n",ans); return 0; }