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.
最简单的AC自动机+DP了。。。。再不会的话AC自动机白学了。
#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;
}
原文地址：https://www.cnblogs.com/JYYHH/p/8280569.html