Codeforces Round #434 D. Polycarp's phone book （字典树）

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本文介绍了一种算法，用于从一系列电话号码中找出每个号码的最短唯一子串，确保输入该子串时能在联系人列表中唯一匹配到该号码。

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
3
123456789
100000000
100123456

output
9
000
01

input
4
123456789
193456789
134567819
934567891

output
2
193
81
91

题意：给你一些电话号码，让你找出每一个电话号码的最短的子串，能唯一的找到这个电话号码，就是找出每个串的最短且不是其他串的子串的子串。

可以用字典树存下每个串的所有后缀（插入的时候把每个后缀的所有前缀的个数都记录），然后查询的时候对于一个串的所有后缀，找出字典树中第一个不存在于其他串中的前缀，然后取长度最短的那个。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

int mx;
char a[70010][10],temp[10],ans[10];

struct trie
{
    trie *next[10];
    int cnt;
    trie()
    {
        memset(next,0,sizeof(next));
        cnt=0;
    }
};
trie *root;
void insert(char *s)
{
    trie *p=root;
    int i,k;
    for(i=0;s[i]!='\0';++i)
    {
        k=s[i]-'0';
        if(p->next[k]==NULL)
            p->next[k]=new trie();
        p=p->next[k];
        p->cnt++;
    }
}

void Delete(char *s)
{
    trie *p = root;
    int i,k;
    for(i=0;s[i]!='\0';i++)
    {
        k = s[i]-'0';
        p = p->next[k];
        p->cnt--;
    }
}
int query(char *s)
{
    trie *p = root;
    int len = 0;
    for(int i=0;s[i]!='\0';i++)
    {
        int x = s[i] - '0';
        temp[len++] = s[i];
        p = p->next[x];
        if(p->cnt == 0)
        {
            if(len < mx)
            {
                mx = len;
                temp[len] = '\0';
                strcpy(ans,temp);
                break;
            }
        }
    }

}
void del(trie *p)
{
    for(int i=0;i<10;i++)
        if(p->next[i]!=NULL)
            del(p->next[i]);
    free(p);
}

int main(void)
{
    int n,i,j;
    while(scanf("%d",&n)==1)
    {
        root=new trie();
        for(i=1;i<=n;i++)
        {
            scanf("%s",a[i]);
            for(j=0;j<9;j++)
                insert(a[i]+j);
        }
        for(i=1;i<=n;i++)
        {
            mx = 10;
            for(j=0;j<9;j++)
                Delete(a[i]+j);
            for(j=0;j<9;j++)
                query(a[i]+j);
            for(j=0;j<9;j++)
                insert(a[i]+j);
            printf("%s\n",ans);
        }

        del(root);
    }

    return 0;
}