Trie樹+簡單DP
對所有的primitives建一個Trie樹。用于查找某字符串是否為primitive。一開始的時候，我圖方便用set來存儲，結果超時。后改成trie。
用dp[i]來保存從i開始的子串的最大prefix數。
這樣如果字符串buf[i..j]是primitive，且dp[i+j]+j>dp[i]，那更新。由于primitive最長為10。只需看前十個字符即可。
時間復雜度為O(10*10*n)。
ps.這題的輸入有點麻煩
#include?<iostream>
#include?<fstream>
#include?<sstream>
#include?<set>
using?namespace?std;
struct?trie_node{
????trie_node*?next[26];
????bool?is_terminal;
????trie_node(){
????????memset(next,0,sizeof(next));
????????is_terminal?=?false;
????}
};
void?insert_str(trie_node*root,const?char*str,int?len)
{
????trie_node?*?cur=root;
????for(int?i=0;i<len;++i){
????????if(?cur->next[str[i]-'A']==NULL){
????????????cur->next[str[i]-'A']?=?new?trie_node;
????????}
????????cur?=?cur->next[str[i]-'A'];?
????}
????cur->is_terminal?=?true;
}
bool?find_str(trie_node*root,const?char?*str,int?len)
{
????trie_node?*?cur?=?root;
????for(int?i=0;i<len;++i){
????????if(?cur->next[str[i]-'A']?==?NULL?)
????????????return?false;
????????else?
????????????cur?=?cur->next[str[i]-'A'];
????}
????if(cur->is_terminal)?
????????return?true;
????else
????????return?false;
}
char?buf[200000];
int?dp[2000001];
int?buf_len;
trie_node?root;
void?input()
{
????freopen("prefix.in","r",stdin);
????freopen("prefix.out","w",stdout);
????while(scanf("%s",buf)&&strcmp(buf,".")!=0){
//????????printf("%s?",buf);
??????????insert_str(&root,buf,strlen(buf));
????}
//????printf("\n");
????int?c;
????buf_len=0;
????while(?(c=getchar())!=EOF){
????????if(!isspace(c)){
????????????buf[buf_len++]?=?(char)c;
//????????????printf("%c",c);
????????}
????}
//????printf("%d\n",buf_len);
}
void?solve()
{
????input();
????dp[buf_len]=0;
????for(int?i=buf_len-1;i>=0;--i){
???????for(int?j=1;j<=10&&i+j-1<buf_len;++j){
???????????if(?find_str(&root,&buf[i],j)?)
???????????????if(dp[i+j]+j>dp[i])
???????????????????dp[i]?=?dp[i+j]+j;
???????}?
????}
????printf("%d\n",dp[0]);
}
int?main(int?argc,char?*argv[])
{
????solve();?
????return?0;
}