Phone List

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

 

1.数据量比较大，申请后要释放。

2.静态的更好一点

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

bool flag;

struct node{
    bool exist;
    node* next[10];
    node(){
        exist=0;
        memset(next,0,sizeof(next));
    }
};

node* rt;
node* root=NULL;

void build(char *a){
    rt=root;
    bool f=0;
    int t=0,len=strlen(a);
    for(int i=0;i<len;i++){
        int id=a[i]-'0';
        if(rt->next[id]) t++;
        if(rt->exist==1) f=1;
        if(!rt->next[id]) rt->next[id]=new node();
        rt=rt->next[id];
    }
    rt->exist=1;
    if(t==len||f) flag=1;
}

void del(node* root){
    for(int i=0;i<10;i++){
        if(root->next[i])
           del(root->next[i]);
    }
    delete(root);
}
int main()
{   int n;scanf("%d",&n);
    while(n--){
        int m;scanf("%d",&m);
        flag=0;
        root=new node();
        for(int i=0;i<m;i++){
            char s[20];
            scanf("%s",s);
            build(s);
        }
        if(flag) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
        del(root);
    }
}