HDU 1247 Hat's Words （字典树）

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【题目链接】click here~~

【题目大意】A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary. ，找出由两个子字符串组成的字符串。

【解题思路】字典树

#include <bits/stdc++.h>using namespace std;const int N=5*1e4+100;const int MOD=998244353;#pragma comment(linker,"/STACK:102400000,102400000")int n,m,k,tot;char word[N][27];typedef  long long LL;struct dictree{    int count;    bool ok;    dictree *tree[27];    dictree()    {        ok=false;        memset(tree,0,sizeof(tree));    }};void insert(dictree *head,char str[]){    dictree *p=head;    int i=0;    while(str[i])    {        int k=(int)(str[i++]-'a');        if(p->tree[k]==NULL) p->tree[k]=new dictree();        p=p->tree[k];    }    p->ok=true;}bool search(dictree *head,char str[]){    int i=0,top=0,num[N];    dictree *p=head;    while(str[i])    {        int k=(int)str[i++]-'a';        if(p->tree[k]==NULL) return false;        p=p->tree[k];        if(p->ok&&str[i]) num[top++]=i;    }    while(top)    {        bool okk=true;        i=num[--top];        p=head;        while(str[i])        {            int k=(int)str[i++]-'a';            if(p->tree[k]==NULL)            {                okk=false;                break;            }            p=p->tree[k];        }        if(okk&&p->ok)  return true;    }    return false;}int main(){    int len=0;    dictree *head=new dictree();    while(gets(word[len]))    {        insert(head,word[len]);        len++;    }    for(int i=0; i<len; ++i)    {        if(search(head,word[i]))            printf("%s\n",word[i]);    }    return 0;}</span>

set容器:

#include <bits/stdc++.h>using namespace std;int main(){    string str;    set<string>sa,sb;    while(cin>>str) sa.insert(str);    set<string>::iterator be=sa.begin(),en=sa.end(),op,opp;    for(; be!=en; be++)    {        int Count=be->size();        for(op=be,op++; op!=en; op++)        {            if(be->compare(op->substr(0,Count))==0)            {                if(sa.count(op->substr(Count,op->size()-Count))) sb.insert(*op);            }            else break;        }    }    for(opp=sb.begin(); opp!=sb.end(); opp++)        cout<<*opp<<endl;    return 0;}</span>

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