Codeforces Round #434 (Div.2)

D. Polycarp's phone book

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n phone numbers in Polycarp's contacts on his phone. Each number is a 9-digit integer, starting with a digit different from 0. All the numbers are distinct.

There is the latest version of Berdroid OS installed on Polycarp's phone. If some number is entered, is shows up all the numbers in the contacts for which there is a substring equal to the entered sequence of digits. For example, is there are three phone numbers in Polycarp's contacts: 123456789, 100000000 and 100123456, then:

• if he enters 00 two numbers will show up: 100000000 and 100123456, 
• if he enters 123 two numbers will show up 123456789 and 100123456, 
• if he enters 01 there will be only one number 100123456. 

For each of the phone numbers in Polycarp's contacts, find the minimum in length sequence of digits such that if Polycarp enters this sequence, Berdroid shows this only phone number.

Input

The first line contains single integer n (1 ≤ n ≤ 70000) — the total number of phone contacts in Polycarp's contacts.

The phone numbers follow, one in each line. Each number is a positive 9-digit integer starting with a digit from 1 to 9. All the numbers are distinct.

Output

Print exactly n lines: the i-th of them should contain the shortest non-empty sequence of digits, such that if Polycarp enters it, the Berdroid OS shows up only the i-th number from the contacts. If there are several such sequences, print any of them.

Examples

input

3123456789100000000100123456

output

900001

input

4123456789193456789134567819934567891

output

21938191

题意:

有n个长度为9的电话号码，找出每个号码的只属于它们的最短子号码。

题解:

字典树，注意末尾。

#include<iostream>#include<string.h>#include<stdio.h>#include<vector>using namespace std;int n,head[15];char num[70000+10][15];struct node{    char num;int cnt,be;    vector<node*>next;    node()    {        num = cnt = be = 0;        next.clear();    }}*tree,*add;int insert_(node *now,int x,int len,int pos){    if(x==len) return now->cnt;    int si = (int) now->next.size();    for(int i=0;i<si;i++)    {        if(now->next[i]->num!=num[pos][x])            continue;        if(now->next[i]->be!=pos)            now->next[i]->be = pos,now->next[i]->cnt++;        return insert_(now->next[i],x+1,len,pos);    }    node *add = new node;    add->be = pos;    add->num = num[pos][x];    add->cnt = 1;    now->next.push_back(add);    return insert_(add,x+1,len,pos);}string find_(node *now,int x,int len,int pos){    string ans = "";    if(x==len) return "";    int si = (int) now->next.size();    for(int i=0;i<si;i++)    {        if(now->next[i]->num!=num[pos][x])            continue;        if(now->next[i]->cnt==1)        {            string s = "";            s+=num[pos][x];            return s;        }        ans += num[pos][x]+find_(now->next[i],x+1,len,pos);        if(x==len-1) return "0000000000";        return ans;    }    return ans;}int main(){    tree = new node;    scanf("%d",&n);    memset(head,-1,sizeof head);    for(int i=1;i<=n;i++)    {        scanf("%s",num[i]);        for(int j=0;j<9;j++)        {            insert_(tree, j, 9, i);        }    }    for(int i=1;i<=n;i++)    {        string s = "";        int len = 111;        for(int j=0;j<9;j++)        {            string ss = find_(tree, j, 9, i);            if(ss.size()<len)            {                s = ss;                len = ss.size();            }        }        cout<<s<<endl;    }    return 0;}