POJ 2001 Shortest Prefixes 【LCP(Trie)】

本文介绍了一种算法,用于从一组字符串中找到每个字符串的最短唯一前缀,通过构建Trie树来高效地解决该问题。适用于词汇缩写、关键字匹配等场景。

Shortest Prefixes
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 17620 Accepted: 7664

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

题意:给字符串的集合,问任何一个字符串的缩写,缩写的定义为:比最大的LCP大1,但如果一个字符串与其他字符串的LCP是他自己,那么它的缩写就是他自己;


思路:构建Trie,从头开始枚举字符串,一位一位比较,直到以这一位为尾的公共前缀的个数为1,那么这个前缀就是这个串的缩写;还有一种情况,搜索到单词结束都不满足条件,那么这时这个串的缩写就是他自己了;


失误:用动态的耗时太大,改成静态的;


AC代码:

#include<cstdio>
#include<cstring>

char S[1100][22];
int sn=0;

struct tree{
	int num;
	int son[26];
}Node[100000];

void init()
{
    int i=0; sn=0;
	for(i=0;i<100000;++i)//结构体中由数组不要用memset()
	{
		int j=0; Node[i].num=0;
		for(j=0;j<26;++j) Node[i].son[j]=0;
	} 
}
void insert(char *str)
{
	int i=0,p1=0;
    while(str[i]) {
    	int pos=str[i]-'a';
    	if(Node[p1].son[pos]==0) {
    		Node[p1].son[pos]=++sn; 
		} 
		p1=Node[p1].son[pos]; ++i;
		Node[p1].num++;
	}
}
int Query(char *str)
{
	int i=0,p=0;
	while(str[i]) {
		int pos=str[i]-'a';
	    p=Node[p].son[pos]; ++i;
	    if(Node[p].num==1) return i; 
	}
	return i;
}
int main()
{
	int cnt=0,i=0; init();
	while(gets(S[++cnt]),strcmp(S[cnt],"")) insert(S[cnt]);
	for(i=1;i<cnt;++i)
	{
		printf("%s ",S[i]);
		int len=Query(S[i]); S[i][len]='\0'; 
		printf("%s\n",S[i]);
	}
	return 0;
}


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