poj 2001 Shortest Prefixes（字典树）

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.

 

题意：给你一系列数字然后让你求他们独一无二的前缀。

 

简单的字典树问题，直接套模版即可

 

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
char str[1010][30];
struct TnT {
    TnT *next[30];
    int v;
    TnT():v(0) {
        memset(next , 0 , sizeof(next));
    }
};
void build(TnT *root , char s[]) {
    int len = strlen(s);
    TnT *p = root;
    for(int i = 0 ; i < len ; i++) {
        int id = s[i] - 'a';
        if(p->next[id] == NULL) {
            p->next[id] = new TnT;
        }
        p = p->next[id];
        p->v++;
        
    }
}
void de(TnT *root) {
    if(root == NULL)
        return ;
    else {
        for(int i = 1 ; i < 27 ; i++) {
            de(root->next[i]);
        }
    }
    delete root;
}
int find(TnT *root , char s[]) {
    int len = strlen(s);
    TnT *p = root;
    for(int i = 0 ; i < len ; i++) {
        int id = s[i] - 'a';
        p = p->next[id];
        if(p->v == 1) {
            return i;
        }
    }
    return len - 1;
}
int main()
{
    TnT *p = new TnT;
    int n = 0;
    while(scanf("%s" , str[n]) != EOF) {
        build(p , str[n++]);
    }
    for(int i = 0 ; i < n ; i++) {
        int gg = find(p , str[i]);
        printf("%s " , str[i]);
        for(int j = 0 ; j <= gg ; j++) {
            printf("%c" , str[i][j]);
        }
        printf("
");
    }
    de(p);
    return 0;
}