目录

1.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

2.代码

#define _CRT_SECURE_NO_WARNINGS   
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 26;
struct Trie
{
  Trie *Next[maxn];  //当前节点可以延伸出的边
  int flag;   
  Trie()  //函数,初始化以该信息为前缀的信息个数
  {
    flag = 1;
    memset(Next, NULL, sizeof(Next));
  }
}*root; //根节点
void insert(char *str)
{
  int len = strlen(str);
  Trie *p = root, *q;
  for (int i = 0; i < len; i++)
  {
    int id = str[i] - 'a';
    if (p->Next[id] == NULL)如果该节点指向的下一个节点为空（没有一条边延伸出去）
    {
      q = new Trie();   //新建一个节点
      p->Next[id] = q;  //存储节点
      p = p->Next[id];  ///指向该节点
    }
    else
    {
      p = p->Next[id];
      ++(p->flag);  ///累加个数
    }
  }
}
int search(char *str)
{
  int len = strlen(str);
  Trie *p = root;
  for (int i = 0; i < len; i++)
  {
    int id = str[i] - 'a';
    p = p->Next[id];
    if (p == NULL) return 0;
  }
  return p->flag;
}
///释放
void free(Trie *T)
{
  if (T == NULL) return;
  for (int i = 0; i < maxn; i++)
  {
    if (T->Next[i])
    {
      free(T->Next[i]);
    }
  }
  delete(T);
}
int main()
{
  char str[20];
  root = new Trie(); //建立根节点
  while (gets(str))
  {
    if (str[0] == '\0') break;
    insert(str);
  }
  while (scanf("%s", str) != EOF)
  {
    printf("%d\n", search(str));
  }
  free(root);
  return 0;
}