HDU 1247 Hat's Words （字典树）

【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>