M - Shortest Prefixes（ｔｒｉｅ树 寻找多个字符各自最短不重复的前缀）

题目

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

解释

将字符串一个个存入字典树，　对于每个访问到的节点node->cont加１。存入后，对每个字符串在字典树中查询，一位一位查如果node->cont不为１说明此处仍然与其它字符存在重叠，继续输出，直到node->cont为１输完这一位后，跳出循环。对于已有abcd, abc的情况下，　ａｂｃ虽然最后一位ｃ的ｃｏｎｔ值不为１。但是输完就结束了，不影响结果。

#include <cstdio>
#include <cstring>
#include <cstdlib>
typedef struct node{
    int cont;
    struct node *next[26];
    int exist;
}trieNode, *Trie;

Trie  buildtrieNode(){
    Trie node = (Trie)malloc(sizeof(trieNode));
    node->cont = 0;
    node->exist = 0;
    memset(node->next, 0, sizeof(node->next));
    return node;
}
void trieNode_insert(char *word, Trie root){
    Trie node = root;
    int id;
    char *p = word;
    while(*p){
        id = *p - 'a';
        if(node->next[id] == NULL){
           node->next[id] = buildtrieNode();
        }
        node = node->next[id];
        p++;
        node->cont++;
    }
    node->exist  = 1;
}
void Trie_search(char *word, Trie root){
    Trie node = root;
    int id;
    char *p = word;

    while(*p){
        id = *p - 'a';
        node = node->next[id];
        printf("%c", *p);
        p++;
        if(node->cont == 1){
            break;
        }
    }
}
int main(){
    int n, i;
    char s[1000][40];
    int cont = 0;
    Trie root = buildtrieNode();
    while(~scanf("%s", s[cont])){
        trieNode_insert(s[cont++], root);
    }
    for(i = 0; i < cont; i++){
        printf("%s ", s[i]);
        Trie_search(s[i], root);
        printf("\n");
    }
    return 0;
}