poj 2001 Shortest Prefixes 【字典树】【找每一个字符串在字符串集里面的 最短且可唯一标识 的前缀】

Shortest Prefixes

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 14747 Accepted: 6364

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

carbohydratecartcarburetorcaramelcariboucarboniccartilagecarboncarriagecartoncarcarbonate

Sample Output

carbohydrate carbohcart cartcarburetor carbucaramel caracaribou caricarbonic carbonicartilage carticarbon carboncarriage carrcarton cartocar carcarbonate carbona

第一道字典树：0ms

#include <cstdio>#include <cstring>#define MAX 10000+10using namespace std;char str[1010][30];int ch[MAX][30];int word[MAX];//记录当前节点下有几个单词 int n = 0;//单词个数 int sz;//节点数 void init(){    sz = 1;//只有一个根节点     memset(ch[0], 0 ,sizeof(ch[0]));    memset(word, 0, sizeof(word));}int idx(char x)//字母编号 {    return x-'a';}void insert(char *s)//插入 {    int i, j, l = strlen(s);    int u = 0;    for(i = 0; i < l; i++)    {        int c = idx(s[i]);        if(!ch[u][c])//节点不存在        {            memset(ch[sz], 0, sizeof(sz));//初始化新节点             ch[u][c] = sz;            sz++; //节点数自增一         }         u = ch[u][c];        word[u]++;//单词数加一     }}void find(char *s)//查询 {    int i, j, l = strlen(s);    int u = 0;//从根节点开始     for(i = 0; i < l; i++)    {        int c = idx(s[i]);        u = ch[u][c];        printf("%c", s[i]);        if(word[u] == 1)//当前前缀足以找出单词         return ;    }}int main(){    init();    while(scanf("%s", str[n]) != EOF)    {        insert(str[n]);//插入         n++;    }    for(int i = 0; i < n; i++)    {        printf("%s ", str[i]);        find(str[i]);        printf("\n");    }    return 0;}