poj2001

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.

【分析】
字典树的模板题。翻译一下大概就是输入一些个单词，然后求出每个单词的前缀（不能有歧义，即前缀不能重复）
建树时，用pos[v]记录v节点的经过次数，若v==1，则说明该前缀很清真，停止输出。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct node{    int v;}pos[50001];string s,sl[5001];int trie[50001][27],sum=1;void insert(string s){    int p=1;    pos[1].v++;    for(int i=0;i<s.size();i++)      if(trie[p][s[i]-'a']!=0)  //p节点的儿砸      {          p=trie[p][s[i]-'a'];          pos[p].v++;      }      else      {          trie[p][s[i]-'a']=++sum;          pos[sum].v++;          p=sum;      }}void find(string s){    for(int i=0;i<s.size();i++) printf("%c",s[i]);    printf(" ");    int p=1;    for(int i=0;i<s.size();i++)    {        printf("%c",s[i]);        if(pos[trie[p][s[i]-'a']].v==1)          break;        p=trie[p][s[i]-'a'];    }    printf("\n");}int main(){    int t=0;    while(cin>>s)    {        sl[++t]=s;        insert(s);    }    for(int i=1;i<=t;i++)      find(sl[i]);    return 0;}