hdu 1247 hat‘s word

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题目链接

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

aahathathatwordhzieeword

Sample Output

ahathatword

题意:给出若干个单词，找出其中能够由其它两个单词连接组合而成的单词。比如上面sample中的"ahat"能够由"a"和"hat"拼凑而成，"hatword"="hat"+"word".因此只需对每个单词进行切分，把切分得到的两个单词在原单词序列中查找，若能查到则符合条件。在这里为了降低查找的复杂度，采用Trie树存储原始单词序列。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;struct node{    node* next[28];    int count;    node():count(0){memset(next,0,sizeof(next));}};char ss[50005][30];node *root=new node;void Insert(char *str){    int i;    node *t=root;    int l=strlen(str);    for(i=0;i<l;i++)    {        if(t->next[str[i]-'a']==0)        {            node *p=new node;            t->next[str[i]-'a']=p;            t=t->next[str[i]-'a'];        }        else        {            t=t->next[str[i]-'a'];        }    }    t->count=1;}int Find(char *str){    node *t=root;    int i,l=strlen(str);    for(i=0;i<l;i++)    {        t=t->next[str[i]-'a'];        if(!t)            return 0;    }    if(t->count)        return 1;    else        return 0;}int main(){    char str[50];    int k=0;    while(scanf("%s",str)!=EOF)    {        Insert(str);        strcpy(ss[k],str);        k++;    }    int i,j,l;    char str1[50],str2[50];    for(i=0;i<k;i++)    {        l=strlen(ss[i]);        for(j=1;j<l;j++)        {            strncpy(str1,ss[i],j);            str1[j]='\0';            strcpy(str2,ss[i]+j);            if(Find(str1)&&Find(str2))            {                printf("%s\n",ss[i]);                break;            }        }    }    return 0;}/*#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct node{    int count;    node *next[28];    node()    {        count=0;        for(int i=0;i<28;i++)            next[i]=NULL;    }};node *root=new node;char ss[50005][30];node *p;node *q;void creatTrie(char *str){    int i,m;    p=root;    int len=strlen(str);    for(i=0;i<m;i++)    {        m=str[i]-'a';        if(p->next[m]!=NULL)        {            p=p->next[m];        }        else        {            q=new node;            p->next[m]=q;            p=q;        }    }    p->count=1;}int find(char *str){    p=root;    int i,m;    int len=strlen(str);    for(i=0;i<len;i++)    {        m=str[i]-'a';        if(p->next[m]==NULL)            return 0;    }        if(p->count)            return 1;        else            return 0;}int main(){    char str[50];    int k=0;    while(scanf("%s",str)!=EOF)    {        creatTrie(str);        strcpy(ss[k],str);        k++;    }    int i,j,l;    char str1[50],str2[50];    for(i=0;i<k;i++)    {        l=strlen(ss[i]);        for(j=1;j<l;j++)        {            strncpy(str1,ss[i],j);            str1[j]='\0';            strcpy(str2,ss[i]+j);            if(find(str1)&&find(str2))            {                printf("%s\n",ss[i]);                break;            }        }    }    return 0;}*/

下面的程序代码存在运行错误，但是我也没改过来，哪位大神呢路过时帮忙给看看哈，不胜感激！

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