hdoj 1247 字典树分词 strncpy函数

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4836    Accepted Submission(s): 1844

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

题目大意：给一组数，找出一种可以是这些数中国两个单词组成的单词

解题思路：先存到字典树里面，然后对每一个单词进行枚举分割判定

注意要点：这个组成的数可能有多重组合，但是只输出一次，多以当输出一次时就要中断，进行下一个。

*恶心之处*：这个题的输入真的很蛋疼，不方便调试

代码：

#include<stdio.h>
#include<malloc.h>
#include<string.h>
char a[50001][51];
struct node
{
    bool has;
    struct node *next[26];
};
void insert(char *str,node *T)
{
    node *p,*q;
    int i,id,j,len;
    p=T;
    len=strlen(str);
    for(i=0;i<len;++i)
    {
        id=str[i]-'a';
        if(p->next[id]==NULL)
        {
            q=(node*)malloc(sizeof(node));
            for(j=0;j<26;++j)
            {
                q->next[j]=NULL;
                q->has=false;
            }
            p->next[id]=q;
        }
        p=p->next[id];
    }
    p->has=true;
    return;
}

int find(char *str,node *T)
{
    int i,id,len,flag;
    node *p;
    p=T;
    len=strlen(str);
    flag=0;
    for(i=0;i<len;++i)
    {
        id=str[i]-'a';
        if(p->next[id]!=NULL)
            p=p->next[id];
        else
            return 0;//这里少个返回，调试了半天
    }
    if(p->has)//判定是否有结尾
        flag=1;
    return flag;
}




int main()
{
    node *T;
    int k,i,len,j;
    char str1[51],str2[51];
    k=0;
    T=(node*)malloc(sizeof(node));
    for(i=0;i<26;++i)
    {
        T->next[i]=NULL;
        T->has=false;
    }
    while(scanf("%s",a[k])!=EOF)
    {
        insert(a[k++],T);
    /*    if(k==6)
            break;*/
    }
    for(i=0;i<k;++i)
    {
        len=strlen(a[i]);
        for(j=0;j<len;++j)
        {
            memset(str1,0,sizeof(str1));
            memset(str2,0,sizeof(str2));
            strncpy(str1,a[i],j+1);
            strcpy(str2,a[i]+j+1);
    //        printf("%s    %s\n",str1,str2);
            if(find(str1,T)&&find(str2,T))
            {
                printf("%s\n",a[i]);
                break;
            }
        }
    }
    return 0;
}