POJ 2001 Shortest Prefixes

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

 

解题思路：使用 trie 树结构。在 trie 树节点中加入两个域count[] 和 next 。count[i] 表示有多少个单词经过这个节点。先将所有单词保存在 trie 树中，然后一个一个地查找，当到达某个节点使用 count[i] ==1 ，那么从根到该节点组成的字符串便是该单词的最短前缀。

#include <stdio.h>
#include <malloc.h>
#include <string.h>

int t = 0;
typedef struct Trie
{
    int count[26];   //统计该字符出现的次数
    struct Trie *next[26];  //26个字母，开辟26个空间
} Trie;


void initTire(Trie *root, char *string)
{
    Trie *trie = root;    //根节点不含有数据，只有26个指针域
    int i;
    int j;

    while(*string != '\0')
    {
        j = *string-'a';
        if(trie->next[j] == NULL)
        {
            trie->next[j] = (Trie *)malloc(sizeof(Trie));
            (trie->next[j])->count[j] = 1;
            trie = trie->next[j];
            for(i = 0; i < 26; i++)
            {
                trie->next[i] = NULL;
            }
        }
        else
        {
            (trie->next[j])->count[j]++;
            trie = trie->next[j];
        }
        string++;
    }
}

void findPrefix(Trie *root, char *string)
{
    //寻找前缀
    Trie *trie = root;
    int i = 0, j;


    while(*string != '\0')
    {
        j = *string-'a';
        if((trie->next[j])->count[j] == 1)
        {
            //如果当前字母只出现一次，证明前缀字母到此结束
            printf("%c", *string);
            break;
        }
        printf("%c", *string);
        trie = trie->next[j];
        string++;
    }
}

int main()
{
    int i = 0;
    int n;
    char str[3000][21];
    char *pre;
    Trie *root;

    root = (Trie *)malloc(sizeof(Trie));
    for(i = 0; i < 26; i++)
    {
        root->next[i] = NULL;
    }

    i = 0;
    while(scanf("%s", str[i]) != EOF)
    {
        initTire(root, str[i]);
        getchar();
        i++;
    }

    n = i;
    for(i = 0; i < n; i++)
    {
        printf("%s ", str[i]);
        findPrefix(root, str[i]);
        printf("\n");
    }
    return 0;
}