HDU 1251

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.

 

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

bananabandbeeabsoluteacmbabbandabc

 

Sample Output

2310

#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);    }}