Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26114    Accepted Submission(s): 8716

 

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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判斷一個有沒有一個串是其他串的字首

插入的時候判斷這個串有沒有出現字首，是不是其他串的字首

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100005
int ch[maxn][10];
int val[maxn];
int sz;
void init()
{
    sz=0;
    //val[0]=0;
    memset(ch,0,sizeof(ch));

}
bool insert(char*s)
{
    int u=0,n=strlen(s);
    bool ok=true;
    for(int i=0;i<n;i++)
    {
        int id=s[i]-'0';

        if(ch[u][id]==0)
    {
        ch[u][id]=++sz;
        val[sz]=0;
        memset(ch[sz],0,sizeof(ch[sz]));

    }
    u=ch[u][id];
    if(val[u]>0)
        ok=false;

}
val[u]=n;
if(ok)
{for(int i=0;i<10;i++)
if(ch[u][i]!=0)
    {ok=false;
break;
    }
}
return ok;
}
char s[20];
int main()
{int t;
scanf("%d",&t);
while(t--)
{
    int n;

scanf("%d",&n);
init();
bool ok=true;
for(int i=0;i<n;i++)
   {
       scanf("%s",s);
       if(ok)
       ok= insert(s);

    }
if(ok)
    printf("YES\n");
else
    printf("NO\n");


}
return 0;
}