通过一棵数来存储每一个位置的字符的个数以及该字符是否出现。
主要代码
Trie *root = NULL;
void Insert(char *word)
{
Trie *cur = root;
int i,len;
len = strlen(word);
for(i = 0; i < len; i++)
{
int id = word[i]-'a';
if(cur->child[id] == NULL)
cur->child[id] = new Trie;
else
cur->child[id]->num++;
cur = cur->child[id];
}
}
统计难题 点击此处
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 ba b band abc
Sample Output
2 3 1 0
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 26;
const int M = 11;
char s[M];
struct Trie
{
int num;
Trie *child[maxn];
Trie()
{
for(int i = 0; i < maxn; i++)
child[i] = NULL;
num = 1;
}
};
Trie *root = NULL;
void Insert(char *word)
{
Trie *cur = root;
int i,len;
len = strlen(word);
for(i = 0; i < len; i++)
{
int id = word[i]-'a';
if(cur->child[id] == NULL)
cur->child[id] = new Trie;
else
cur->child[id]->num++;
cur = cur->child[id];
}
}
int F(char *word)
{
Trie *cur = root;
int i,len;
len = strlen(word);
for(i = 0; i < len; i++)
{
int id = word[i]-'a';
cur = cur->child[id];
if(cur == NULL)
return 0;
}
return cur->num;
}
int main()
{
root = new Trie;
while(gets(s) && s[0]!='\0')
{
Insert(s);
}
while(scanf("%s",s)!=EOF)
{
printf("%d\n",F(s));
}
return 0;
}
POJ 2001 点击此处
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 26;
const int N = 1000+10;
const int M = 21;
char dic[N][M];
struct Trie
{
int childNum;
Trie *child[maxn];
Trie()
{
for(int i = 0; i < maxn; ++i)
child[i] = NULL;
childNum = 1;
}
};
Trie *root = NULL;
void Insert(char *word)
{
Trie *cur = root;
int len = strlen(word);
for(int i = 0; i < len; ++i)
{
int id = word[i]-'a';
if(cur->child[id] == NULL)//下继存在就开一个空间
cur->child[id] = new Trie;
else
cur->child[id]->childNum++;
cur = cur->child[id];
}
}
int findshortest(char *word)
{
int i,len = strlen(word);
Trie *cur = root;
for(i = 0; i < len; ++i)
{
int id = word[i]-'a';
cur = cur->child[id];
if(cur->childNum <= 1)
return i;
}
return -1;
}
int main()
{
int ant = 0;
root = new Trie;
while(scanf("%s",dic[ant])!=EOF)
Insert(dic[ant++]);
for(int j = 0; j < ant; j++)
{
int ans = findshortest(dic[j]);
if(ans == -1)
printf("%s %s\n",dic[j],dic[j]);
else
{
printf("%s ",dic[j]);
dic[j][ans+1] = '\0';
printf("%s\n",dic[j]);
}
}
return 0;
}