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
3 7
ABA
CB
ABACB
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.
题解：
记cnt[i]为节点i沿着fail一直走下去可以获得的积分,那么
f[i][j]为走了i步到节点j的最大积分 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);
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
const int INF=-2e8;
struct node
{
int next[];
int cnt;
}a[];
int f[][];
int root=,num=,fail[];
char s[];
void Clear()
{
a[num].cnt=;
for(int i=;i<;i++)a[num].next[i]=;
}
void add()
{
scanf("%s",s);
int p=root;
for(int i=,ls=strlen(s);i<ls;i++)
{
if(a[p].next[s[i]-'A'])p=a[p].next[s[i]-'A'];
else
{
a[p].next[s[i]-'A']=++num;
Clear();
p=num;
}
}
a[p].cnt++;
}
void getfail()
{
queue<int>q;
q.push(root);
int u,p,v;
while(!q.empty())
{
u=q.front();q.pop();
for(int i=;i<;i++)
{
if(!a[u].next[i])
{
if(a[fail[u]].next[i])a[u].next[i]=a[fail[u]].next[i];
continue;
}
p=fail[u];
while(p)
{
if(a[p].next[i])break;
p=fail[p];
}
if(a[p].next[i] && a[p].next[i]!=a[u].next[i])fail[a[u].next[i]]=a[p].next[i];
v=a[u].next[i];
a[v].cnt+=a[fail[v]].cnt;
q.push(a[u].next[i]);
}
}
}
int main()
{
//freopen("pp.in","r",stdin);
int n,k,ans=;
scanf("%d%d",&n,&k);
for(int i=;i<=n;i++)
add();
getfail();
for(int i=;i<=k;i++)
for(int j=;j<=num;j++)
f[i][j]=INF;
f[][]=;
for(int i=;i<=k;i++)
{
for(int j=;j<=num;j++)
{
for(int k=;k<;k++)
{
f[i][a[j].next[k]]=max(f[i][a[j].next[k]],f[i-][j]+a[a[j].next[k]].cnt);
}
}
}
for(int i=;i<=num;i++)if(f[k][i]>ans)ans=f[k][i];
printf("%d\n",ans);
return ;
}