Phone List

Posted 2020-08-16 勿忘初心0924

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18386    Accepted Submission(s): 6192

Problem 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:
1. Emergency 911
2. Alice 97 625 999
3. 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

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

 

 

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/*
如果一个电话号码是另一个电话号码的前缀的话，那么就会产生错误

在字典树上加一个v数组表示以i节点为前缀的单词数
如果一个单词在树上是两个单词或以上前缀的话，就说明他除了自己还是别人的前缀就产生错误了
*/
#include<bits/stdc++.h>
using namespace std;
#define MAX 26
#define N 100005
#define M 100
const int maxnode=4000*100+100;///预计字典树最大节点数目
const int sigma_size=10;///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-‘0‘;
    }

    ///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 false;
            u=ch[u][id];
            //cout<<"s="<<s<<" v[u]="<<v[u]<<endl;
        }
        return v[u];
    }
};
Trie trie;///定义一个字典树
char op[N][M];
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        trie.Clear();
        scanf("%d",&n);
        getchar();
        int f=0;
        for(int i=0;i<n;i++)
        {
            //cout<<i<<endl;
            scanf("%s",op[i]);
            //getchar();
            //cout<<"=="<<op[i]<<"=="<<endl;
            trie.Insert(op[i]);
        }
        for(int i=0;i<n;i++)
        {
            //cout<<"trie.Search(op[i])="<<trie.Search(op[i])<<endl;
            //cout<<"op[i]="<<op[i]<<endl;
            if(trie.Search(op[i])>=2)
            {
                f=1;
                break;
            }
        }
        if(f)
            puts("NO");
        else
            puts("YES");
    }
    return 0;
}