hdu 1247

Problem Description

A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
You are to find all the hat’s words in a dictionary.

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
Only one case.

Output

Your output should contain all the hat’s words, one per line, in alphabetical order.

Sample Input

a

ahat

hat

hatword

hziee

word

Sample Output

ahat

hatword

题解：刚开始就老想着在  find函数里修改，其实  模板不用变，就是多了对字符串的处理。

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <ctype.h>
#include <iomanip>
#include <queue>
#include <stdlib.h>
using namespace std;

struct Tri
{
    bool v;
    Tri* child[26];
};

Tri* root;

void Init()
{
    root->v=false;
    for(int i=0;i<26;i++)
    {
        root->child[i]=NULL;
    }
}

void CreateDic(char* s)
{
    Tri* p;
    int j;
    int len=strlen(s);
    if(len==0)
        return ;
    p=root;
    for(int i=0 ;i < len; i++)
    {
        if(p->child[s[i]-'a']==NULL)
        {
            p->child[s[i]-'a']=(Tri*)new Tri;
            p->child[s[i]-'a']->v=false;
            for(j=0;j<26;j++)
                p->child[s[i]-'a']->child[j]=NULL;
        }
        p=p->child[s[i]-'a'];
        
    }
    p->v=true;
}

bool Find(char *s)
{
    Tri* p=root;
    int len=strlen(s);
    if(len==0)
        return 0;
    for(int i=0 ;i < len; i++)
    {
        if(p->child[s[i]-'a']==NULL)
            return 0;
        p=p->child[s[i]-'a'];
    }
    return p->v;
}

void Del(Tri* p)
{
    for(int i=0;i<26;i++)
        if(p->child[i])
            Del(p->child[i]);
    Del(p);
    
}

char total[50002][100];
int main()
{
    char a[100],b[100],c[100];
    int i=0,j,k;

    root=(Tri*)new Tri;
    Init();
    while(gets(a))
    {
        CreateDic(a);
        strcpy(total[i++],a);
    }

    for(j=0;j<i;j++)
    {
        for(k=1;k<strlen(total[j]);k++)
        {
            strncpy(b,total[j],k);
            b[k]='\0';
            strcpy(c,total[j]+k);
                
            if(Find(b)&&Find(c))
            {
                cout<<total[j]<<endl;
                break;   //注意结束循环                   
            }
        }
    }
    return 0;
}