hdu1247
Hat’s Words
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6422Accepted 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.
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.
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,'\0',sizeof(a) );
memset( b,'\0',sizeof(b) );
strncpy(a,st[i],j);
strncpy(b,st[i]+j,len-j);
if(find(a)==1&&find(b)==1)
{
printf("%s\n",st[i]);
break;
}
}
}
return 0;
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#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,'\0',sizeof(a) );
memset( b,'\0',sizeof(b) );
strncpy(a,st[i],j);
strncpy(b,st[i]+j,len-j);
if(find(a)==1&&find(b)==1)
{
printf("%s\n",st[i]);
break;
}
}
}
return 0;
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