hdu 1671 Phone List 统计前缀次数

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27948    Accepted Submission(s): 9204

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

 

题意：判断电话号码是不是其它电话号码的前缀，是输出YES,否则输出NO

 

注意开数组的大小100050

#include<iostream>
#include<string>
#include<string.h>
using namespace std;
int vis[],tree[][];
int root,id,len,n,t,i,num,flag;
string s[];

void insert()
{
    root=;
    len=s[i].length();
    for(int j=;j<len;j++)
    {
        id=s[i][j]-'a';
        if(!tree[root][id])
            tree[root][id]=++num;
        vis[tree[root][id]]++;
        root=tree[root][id];
    }
}

int search(string ss)
{
    root=;
    len=ss.length();
    for(int j=;j<len;j++)
    {
        id=ss[j]-'a';
        if(!tree[root][id])
            return ;
        root=tree[root][id];
    }
    return vis[root];
}
int main()
{
    cin>>t;
    while(t--)
    {
        flag=,num=,i=;
        memset(tree,,sizeof(tree));
        memset(vis,,sizeof(vis));
        cin>>n;
        for(i=;i<n;i++)
        {
            cin>>s[i];
            insert();
        }
        for(int j=;j<n;j++)
        {
            if(search(s[j])!=)
            {
                flag=;
                cout<<"NO"<<endl;
                break;
            }
        }
        if(flag==)
            cout<<"YES"<<endl;

    }
    return ;
}