Shortest Prefixes

Shortest Prefixes

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 17715		Accepted: 7698

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

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

Source

Rocky Mountain 2004

/*
每插入一个单词，把经过的路径都在v数组中加一
遍历的时候分割单词，如果某一个前缀的最后一个字母只被标记过一次，那么这个前缀就是是能用的
*/
#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#define MAX 26
#define N 25
#define M 1005
using namespace std;
const int maxnode=4000*100+100;///预计字典树最大节点数目
const int sigma_size=26;///每个节点的最多儿子数

struct Trie
{
    ///这里ch用vector<26元素的数组> ch；实现的话，可以做到动态内存
    int ch[maxnode][sigma_size];///ch[i][j]==k表示第i个节点的第j个儿子是节点k
    int val[maxnode];///val[i]==x表示第i个节点的权值为x
    int sz;///字典树一共有sz个节点,从0到sz-1标号
    int v[maxnode];

    ///初始化
    void Clear()
    {
        sz=1;
        memset(ch[0],0,sizeof(ch[0]));///ch值为0表示没有儿子
        memset(v,0,sizeof v);
    }

    ///返回字符c应该对应的儿子编号
    int idx(char c)
    {
        return c-'a';
    }

    ///在字典树中插入单词s,但是如果已经存在s单词会重复插入且覆盖权值
    ///所以执行Insert前需要判断一下是否已经存在s单词了
    void Insert(char *s)
    {
        int u=0,n=strlen(s);
        for(int i=0;i<n;i++)
        {
            int id=idx(s[i]);
            if(ch[u][id]==0)///无该儿子
            {
                ch[u][id]=sz;
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz++]=0;
            }
            u=ch[u][id];
            v[u]++;
        }
        val[u]=n;
    }

    ///在字典树中查找单词s
    int Search(char *s)
    {
        int n=strlen(s),u=0;
        for(int i=0;i<n;i++)
        {
            int id=idx(s[i]);
            if(ch[u][id]==0)
                return -1;
            u=ch[u][id];
        }
        return v[u];
    }
};
char op[M][N];
Trie trie;///定义一个字典树
int main()
{
    //freopen("C:\Users\acer\Desktop\in.txt","r",stdin);
    trie.Clear();
    int len=0;
    while(gets(op[len++]))
    {
        trie.Insert(op[len-1]);
        //cout<<op[len-1]<<endl;
    }
    for(int i=0;i<len-1;i++)
    {
        printf("%s ",op[i]);
        int f=1;
        int n=strlen(op[i]);
        //cout<<"n="<<n<<endl;
        for(int k=1;k<=n;k++)
        {
            char s1[N];
            strncpy(s1,op[i],n);
            s1[k]='';
            //cout<<s1<<endl;
            if(trie.Search(s1)==1)
            {
                f=0;
                puts(s1);
                break;
            }
        }
        if(f)
            puts(op[i]);
    }
    return 0;
}