Shortest Prefixes（字典树）

Shortest Prefixes

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 14879 Accepted: 6424

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.

Sample Input

carbohydratecartcarburetorcaramelcariboucarboniccartilagecarboncarriagecartoncarcarbonate

Sample Output

carbohydrate carbohcart cartcarburetor carbucaramel caracaribou caricarbonic carbonicartilage carticarbon carboncarriage carrcarton cartocar carcarbonate carbona

Source

Rocky Mountain 2004

这一题是求单一前缀的！我们这样做把每一单词的每一个字符的访问度先清为0，每次

插入一个单词的时候，这个单词的每一个字符的访问度加一，搜索的时候每一个单词

的是否有度数为1？为1说明这个字符以及前面的是这个单词的前缀！

像abc abfhj abcd我们看到C的访问度是2那么abc就不是唯一了吗？是的他不为1但是他

可以作为abc的前缀，因为算到c的时候abc已经输出完毕了！

#include<cstdio>#include<iostream>#include<vector>#include<cstring>using namespace std;struct Tree{    int num;//判断此节点是否为空的（就是一棵树的结束点）    Tree *next[26];//这个是这棵树的子节点    Tree()    {        num=0;        for(int i=0;i<26;i++)        {            next[i]=NULL;        }    }//这是一棵树}*root;//建立一个树根int count1;void insert(Tree *p,char *s){    int i=0;    while(s[i])//当这个字符串的某一个元素不为空的时候    {        int x=s[i]-'a';//看这个元素在哪里？        if(p->next[x]==NULL)//然后看这棵树有没有这个元素如果没有        {            p->next[x]=new Tree();//就在这棵树的此节点重新建立一棵子树        }        p=p->next[x];//p指向他的子树，这一点你就可以知道了所有字符串的元素原来他们都是主仆关系没有相同的等级关系！        p->num++;  //记录该节点的访问量（重点啊啊啊啊啊啊啊啊啊啊啊。。。。。。。。。。。。。。。。。。。。）        i++;//继续遍历    }}//插入字符void find(Tree *p,char *s){    int i=0,ans=0;//寻找树的元素，ans是结果！    while(s[i])//当他不为空的时候就是字符串还有的时候    {        int n=s[i]-'a';//找到他在这个树的节点        p=p->next[n];        cout<<s[i];        if(p->num==1)        {            //cout<<"over啦"<<endl;            break;        }        i++;    }    cout<<endl;}int main(){    root=new Tree();    char s[25];    char word[1005][25];    int i=0,j;    while(cin>>word[i])    {        insert(root,word[i]);        i++;    }    for(j=0;j<i;j++)    {        cout<<word[j]<<" ";        find(root,word[j]);    }}