POJ 2001 (tire)

Shortest Prefixes

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8118		Accepted: 3445

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. 

今天看了看TIRE的模板，順手改了改把這題A了。。。。

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

#define max 26
#define maxn1 1001
#define maxn2 30

char as[maxn1][maxn2],tt[maxn2];


struct node//初始tire樹 
{
  int num,k;//標記是否爲相同前綴最後字符 
  node *next[max];//後繼結點 
} * root;

void insert(char ss[])//插入操作 
{
  int i,j;
  node *p=root,*t;//p指向當前節點，t用來開闢內存 
  for(i=0;ss[i];i++)//將每個字符存起來的過程 
  {
    if(p->next[ss[i]-'a']==NULL)//當p指向爲空的時候，需要開闢新的節點 
    {
      t=new node;
      t->k = t->num = 0;
      for(j=0;j<max;j++)
        t->next[j]=NULL;
      p->next[ss[i]-'a']=t;//接上新的節點 
    }
    p->k++;
    p=p->next[ss[i]-'a'];//p指向下一個節點 
    p->num++;//
  }
}

void search(char ss[])//查找 
{
  node *p=root;//p指向根節點 
  int t=0;
  for(int i=0;ss[i];i++)
  {
    p=p->next[ss[i]-'a'];//p指向下一結點 
    if(p->num>1)
    {
      tt[t]=ss[i];
      t++;
    }
    if(p->num==1)
      {
      tt[t]=ss[i];
      tt[t+1]='\0';
      return;
      }
    }
  if(p->k == 1) tt[t]='\0';
  tt[t]='\0';//這個是防止上面的p->num都是大於1的 
  return ;
}

int main()
{ 
    int l = 0,i,j;
    char s[maxn2];
    root = new node;
    root->k = root->num = 0;
      for(j=0;j<max;j++)
        root->next[j]=NULL;
    while(scanf("%s",s) != EOF) {
      //if (s[0] == '0') break;
      strcpy(as[l++],s);
      
      insert(s);
      }
    for(i = 0;i < l;i++) {
      search(as[i]);
      printf("%s %s\n",as[i],tt);
      }
    //scanf("%d",&i);
    return 0;
}