Phone List
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12684 Accepted Submission(s): 4307
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.
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.
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
YES
Source
/**
题意：给出n个电话号码，要求是否存在有的号码是另外一个字符串的前缀
做法：Trie树 一直是RE 是把主函数的root = new node() 写成 p = new node()
然后 C++MLE 然后每次询问完删除树就OK
**/
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;
char ch[][];
struct node
{
int count;
node *next[];
node()
{
count = ;
for(int i=;i<;i++)
{
next[i] = NULL;
}
}
};
node *root = new node(),*p;
void insert(char *s)
{
p = root;
for(int i=;i<strlen(s);i++)
{
int tt = s[i] -'';
if(p->next[tt] == NULL) p->next[tt] = new node();
p = p->next[tt];
p->count++;
}
}
int find(char *c)
{
int i;
p = root;
int len = strlen(c);
for(i=;i<len;i++)
{
int tt = c[i]-'';
if(p->next[tt]->count== ) break;
p = p->next[tt];
}
if(i == len) return ;
else return ;
}
void freedom(node *pp)
{
for(int i=;i<;i++)
{
if(pp->next[i] != NULL)
freedom(pp->next[i]);
}
free(pp);
}
int main()
{
// freopen("in.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--)
{
root = new node();
int n;
bool flag = true;
scanf("%d",&n);
for(int i=;i<n;i++)
{
scanf("%s",ch[i]);
insert(ch[i]);
}
for(int i=;i<n;i++)
{
if(find(ch[i]) == )
{
flag = false;
break;
}
}
if(!flag) printf("NO\n");
else printf("YES\n");
freedom(root);
}
return ;
}