统计难题
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)Total Submission(s): 10040 Accepted Submission(s): 4074
Problem Description
Ignatius最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的前缀).
Input
输入数据的第一部分是一张单词表,每行一个单词,单词的长度不超过10,它们代表的是老师交给Ignatius统计的单词,一个空行代表单词表的结束.第二部分是一连串的提问,每行一个提问,每个提问都是一个字符串.
注意:本题只有一组测试数据,处理到文件结束.
注意:本题只有一组测试数据,处理到文件结束.
Output
对于每个提问,给出以该字符串为前缀的单词的数量.
Sample Input
banana band bee absolute acm ba b band abc
Sample Output
2 3 1 0
#include "stdio.h"
#include "string.h"
#include "stdlib.h"
struct dictree
{
struct dictree *child[26];
int n;
};
struct dictree *root;
void insert (char *source)
{
int len,i,j;
struct dictree *current,*newnode;
len=strlen(source);
if(len==0) return;
current= root;
for(i=0;i<len;i++)
{
if(current->child[source[i]-'a']!=0)
{
current=current->child[source[i]-'a'];
current->n=current->n+1;
}
else
{
newnode=(struct dictree *)malloc(sizeof(struct dictree));
for(j=0;j<26;j++)
newnode->child[j]=0;
current->child[source[i]-'a']=newnode;
current=newnode;
current->n=1;
}
}
}
int find(char *source)
{
int i,len;
struct dictree *current;
len=strlen(source);
if(len==0) return 0;
current=root;
for(i=0;i<len;i++)
{
if(current->child[source[i]-'a']!=0)
current=current->child[source[i]-'a'];
else return 0;
}
return current->n;
}
int main()
{
char temp[11];
int i,j;
root=(struct dictree *)malloc(sizeof(struct dictree));
for(i=0;i<26;i++)
root->child[i]=0;
root->n=2;
while(gets(temp),strcmp(temp,"")!=0)
insert(temp);
while(scanf("%s",temp)!=EOF)
{
i=find(temp);
printf("%d\n",i);
}
}
