学步园

trie树+dp，dp属于很简单的状态dp  dp[i]=dp[i-a[k].len] | dp[i],

code

/*
ID: yueqiq
LANG: C++
TASK: prefix
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ls rt<<1
#define rs rt<<1|1
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("prefix.in","r",stdin)
#define writef freopen("prefix.out","w",stdout)
const double pi = acos(-1.0);
const int maxn = 50;
const int BigP = 99999999;
const int INF = 99999999;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
using namespace std;
char str[200001],str2[80];
bool dp[200001];
struct trie{
    trie *next[26];
    bool is_terminal;
    trie(){
        SET(next,0);
        is_terminal=false;
    }
}root;
void insert(trie *root,char *word){//&root
    int len=strlen(word);
    trie *p=root;
    FF(i,len){
        if(p->next[word[i]-'A']==NULL)
            p->next[word[i]-'A']=new trie;
        p=p->next[word[i]-'A'];
    }
    p->is_terminal=true;
}
bool query(trie *root,char *word,int len){//len为当前位置向后多少位
    trie *p=root;
    FF(i,len){
        if(p->next[word[i]-'A']==NULL) return false;
        p=p->next[word[i]-'A'];
    }
    if(p->is_terminal) return true;
    return false;
}
int main (){
    readf;
    writef;
    while(cin>>str2&&str2[0]!='.') insert(&root,str2);
    while(cin>>str2) strcat(str,str2);
    //printf("%s\n",str);
    int len=strlen(str);
    int ans=0;
    dp[0]=true;
    FF(i,len){
        if(!dp[i]) continue;
        for(int j=1;j<=10&&i+j-1<len;j++){
            if(query(&root,&str[i],j)){
                dp[i+j]=1;
                if(i+j>ans) ans=i+j;
            }
        }
    }
    PD(ans);LN;
	return 0;
}