【Trie】poj 2001 Shortest Prefixes

该博客介绍了一种利用Trie数据结构解决寻找字符串列表中每个单词最短唯一前缀的方法。通过建立Trie树并遍历每个单词,可以找到能唯一标识每个单词的最短前缀,避免歧义。代码示例展示了如何实现这个过程,并给出了输入输出示例。

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Shortest Prefixes

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 27219 Accepted: 11388

Description

A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "car", "carb", "carbo", and "carbon". Note that the empty string is not considered a prefix in this problem, but every non-empty string is considered to be a prefix of itself. In everyday language, we tend to abbreviate words by prefixes. For example, "carbohydrate" is commonly abbreviated by "carb". In this problem, given a set of words, you will find for each word the shortest prefix that uniquely identifies the word it represents.

In the sample input below, "carbohydrate" can be abbreviated to "carboh", but it cannot be abbreviated to "carbo" (or anything shorter) because there are other words in the list that begin with "carbo".

An exact match will override a prefix match. For example, the prefix "car" matches the given word "car" exactly. Therefore, it is understood without ambiguity that "car" is an abbreviation for "car" , not for "carriage" or any of the other words in the list that begins with "car".

Input

The input contains at least two, but no more than 1000 lines. Each line contains one word consisting of 1 to 20 lower case letters.

Output

The output contains the same number of lines as the input. Each line of the output contains the word from the corresponding line of the input, followed by one blank space, and the shortest prefix that uniquely (without ambiguity) identifies this word.

Sample Input

carbohydrate
cart
carburetor
caramel
caribou
carbonic
cartilage
carbon
carriage
carton
car
carbonate

Sample Output

carbohydrate carboh
cart cart
carburetor carbu
caramel cara
caribou cari
carbonic carboni
cartilage carti
carbon carbon
carriage carr
carton carto
car car
carbonate carbona

 

Trie模板题,将所有的单词先插入到Trie树里(记录一下每个点出现的次数),然后搜每个单词,搜到权值为1的点就可以退出了

 

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
int ch[20000+5][27];
int val[20000+5],cnt,size;
char inin[1002][2000+5];
void insert(char *ss)
{
	int now=0,n=strlen(ss);
	for(int i=0;i<n;i++)
	{
		int c=ss[i]-64;
		if(!ch[now][c])
			ch[now][c]=++size; 
		++val[ch[now][c]];
		now=ch[now][c];	
	}
}
void find(char *ss)
{
	int now=0,n=strlen(ss);
	for(int j=0;j<n;j++)
	{
		printf("%c",ss[j]); 
		int c=ss[j]-64;
		now=ch[now][c];	
		if(val[now]==1)
			break;
	}
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	while(scanf("%s",inin[cnt])!=EOF)
	{
		insert(inin[cnt]);
		cnt++;
	}
	for(int i=0;i<cnt;i++)
	{
		int lenn=strlen(inin[i]);
		for(int j=0;j<lenn;j++)
			printf("%c",inin[i][j]);
		printf(" ");
		find(inin[i]);
		printf("\n");
	} 
	return 0;	
} 

 

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