hdu1247

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6422    Accepted Submission(s): 2399

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

 

思路：把每个单词分为两部分，判断两部分是否都在已经建好的字典树中

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<malloc.h>
#define MAX 50010
using namespace std;
char st[MAX][105];
struct tree
{
    struct tree *child[26];
    int v;
}*root;
void init()
{
    int i;
    root=(struct tree*)malloc(sizeof(struct tree));
    for(i=0; i<26; i++)
    {
        root->child[i]=0;
        root->v=0;
    }
    return ;
}
void build(char *str)
{
    int id,i,j,len;
    len=strlen(str);
    struct tree *p=root,*q;
    for(i=0; i<len; i++)
    {
        id=str[i]-'a';
        if(p->child[id]==0)
        {
            q=(struct tree*)malloc(sizeof(struct tree));
            for(j=0; j<26; j++)
                q->child[j]=0;
            p->child[id]=q;
            p=p->child[id];
            p->v=-1;
        }
        else
        {
            p=p->child[id];
        }
    }
    p->v=1;
    return ;
}
int find(char *str)
{
    int len=strlen(str);
    int i,j,id;
    struct tree *p=root;
    for(i=0; i<len; i++)
    {
        id=str[i]-'a';
        if(p->child[id]==0)
            return -1;
        p=p->child[id];
    }
    if(p->v==1)
        return 1;
    else
        return -1;
}
int main()
{
    int cnt,i,j,len;
    char a[105],b[105];
    init();
    cnt=0;
    while(scanf("%s",st[cnt])!=EOF)
    {
        build(st[cnt]);
        cnt++;
    }
    for(i=0; i<cnt; i++)
    {
        len=strlen(st[i]);
        for(j=0; j<len; j++)
        {
            memset( a,'',sizeof(a) );
            memset( b,'',sizeof(b) );
            strncpy(a,st[i],j);
            strncpy(b,st[i]+j,len-j);
            if(find(a)==1&&find(b)==1)
            {
                printf("%s ",st[i]);
                break;
            }
        }
    }
    return 0;
}