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.
Output
For each test case, output "YES" if the list is consistent, or "NO" otherwise.
Sample Output
NO
YES
这道题就是判断一组字符串中是否有一个字符串是另一个字符串的前缀.
#include <iostream>
#include <string>
#include <cstring>
using namespace std;
int k;
typedef structnode
{
int num;
structnode *next[10];
}node;
nodememory[1000000];
int insert(char *s, node*T)
{
int len = 0;
int id = 0;
node*p,*q;
p = T;
len = strlen(s);
for(int i = 0 ; i < len; i++)
{
id = s[i] - '0';
if(p->num == 1)
return 1;
if(p->next[id] == NULL)
{
q = &memory[k++];
q->num = 0;
for(int j = 0; j < 10; j++)
{
q->next[j] = NULL;
}
p->next[id] = q;
}
p = p->next[id];
}
for(int i = 0 ; i < 10; i++)
{
if(p->next[i] != NULL)
return 1;
}
p->num = 1;
return 0;
}
int main()
{
int t;
node*T;
cin>>t;
while(t)
{
k = 0;
T = &memory[k++];
T->num = 0;
for(int i = 0 ; i < 10; i++)
{
T->next[i] = NULL;
}
int flag = 0;
int n;
cin>>n;
for(int i = 0 ; i < n ; i++)
{
char s[15];
cin>>s;
if(flag)
continue;
if(insert(s,T))
flag = 1;
}
if(flag)
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
t--;
}
return 0;
}
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