统计难题
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 15031 Accepted Submission(s): 6436
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
banana
band
bee
absolute
acm
band
bee
absolute
acm
ba
b
band
abc
Sample Output
2 3 1 0
思路：字典树
AC代码：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
typedef struct Tree_Node
{
int cnt;
struct Tree_Node *child[];
}Node;
Node *root;
char str[];
void insert()
{
int i;
if(str == NULL)
return ;
char *p = str;
Node *t = root;
while(*p != '\0')
{
if(NULL == t -> child[*p- 'a'])
{
Node *temp = (Node *)malloc(sizeof(Node));
memset(temp, , sizeof(Node));
for(i = ;i < ;i ++)
{
temp -> child[i] = NULL;
temp -> cnt = ;
}
t -> child[*p - 'a'] = temp;
}
t = t -> child[*p - 'a'];
t -> cnt ++;
p ++;
}
}
int search()
{
if(NULL == root)
return ;
char *p = str;
Node *t = root;
while(*p != '\0')
{
if(NULL != t -> child[*p - 'a'])
{
t = t -> child[*p - 'a'];
p ++;
}
else
return ;
}
return t -> cnt;
} int main(int argc, char const *argv[])
{
int i;
char c[] = "abc";
Node TREEROOT;
root = &TREEROOT;
for(i = ;i < ;i ++)
{
root -> child[i] = NULL;
root -> cnt = ;
}
// freopen("in.c","r",stdin);
while(gets(str) && strcmp(str,""))
{
insert();
memset(str, , sizeof(str));
}
memset(str, , sizeof(str));
while(~scanf("%s", str))
{
printf("%d\n", search());
memset(str, , sizeof(str));
}
return ;
}