zoukankan      html  css  js  c++  java
  • Phone List

    Phone List
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 14044   Accepted: 4529

    Description

    Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers:

    • Emergency 911
    • Alice 97 625 999
    • Bob 91 12 54 26

    In this case, it's not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob's phone number. So this list would not be consistent.

    Input

    The first line of input gives a single integer, 1 ≤ t ≤ 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 ≤ n ≤ 10000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.

    Output

    For each test case, output "YES" if the list is consistent, or "NO" otherwise.

    Sample Input

    2
    3
    911
    97625999
    91125426
    5
    113
    12340
    123440
    12345
    98346

    Sample Output

    NO
    YES

    Source

    #include <iostream>
    #include <fstream>
    #include <vector>
    #include <string>
    #include <algorithm>
    #include <map>
    #include <stack>
    #include <cmath>
    #include <queue>
    #include <set>
    #include <list>
    #include <string.h>
    #include <cstdlib>
    #define REP(i,j,k) for( int i = j ; i < k ; ++i)
    #define MAX (100010)


    using namespace std;


    struct Node
    {
    bool isExist;
    int branch[10];

    void clear()
    {
    this->isExist = false;
    REP(i,0,10)
    {
    this->branch[i] = -1;
    }
    }
    };


    Node buff[MAX];
    int size = 0;


    class Trie
    {
    public:
    int root;

    bool insert( char num[] )
    {
    int location = root;
    int i = 0;
    int len = strlen(num);

    for( ; i < len - 1 ; ++i )
    {
    if(buff[location].isExist == true)
    {
    return false;
    }

    if( buff[location].branch[num[i] - '0'] == -1 )
    {
    buff[location].branch[num[i] - '0'] = ++size;
    buff[size].clear();
    }

    location = buff[location].branch[num[i] - '0'];
    }


    if(buff[location].isExist == true)
    {
    return false;
    }

    if(buff[location].branch[num[i] - '0'] == -1 )
    {
    buff[location].branch[num[i] - '0'] = ++size;
    buff[size].clear();
    buff[size].isExist = true;
    }
    else
    {
    return false;
    }

    return true;
    }
    };




    int main()
    {



    int num_case;
    cin >> num_case;

    Trie trie;




    REP(i,0,num_case)
    {
    int num_phone;
    cin >> num_phone;
    bool tag = false;
    size = 0;
    buff[0].clear();
    trie.root = 0;

    REP(j,0,num_phone)
    {
    char phone[11];

    scanf("%s" , phone);

    if( !tag )
    {
    if ( trie.insert(phone) == false)
    {
    tag = true;
    }
    }


    }

    if(!tag)
    {
    cout << "YES" << endl;
    }
    else
    {
    cout << "NO" << endl;
    }
    }

    return 0;

    }
  • 相关阅读:
    「转」xtrabackup新版详细说明
    微博MySQL优化之路--dockone微信群分享
    分享的好处
    DBA的技能图谱
    高效运维--数据库坐而论道活动
    MySQL的诡异同步问题-重复执行一条relay-log
    把信送给加西亚读后感
    一次由于字符集问题引发的MySQL主从同步不一致问题追查
    nginx解决浏览器跨域问题
    kubernetes之pod调度
  • 原文地址:https://www.cnblogs.com/kking/p/2335270.html
Copyright © 2011-2022 走看看