Phone List
時間限制: 1 Sec 內存限制: 64 MB題目描述
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.
輸入
The first line of input gives a single integer, 1 ≤ t ≤ 10, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 ≤ n ≤ 100000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.
輸出
For each test case, output "YES" if the list is consistent, or "NO" otherwise.
樣例輸入
2
3
911
97625999
91125426
5
113
12340
123440
12345
98346樣例輸出
NO
YES解題思路：典型的字典樹（容易超內存）
代碼如下：
# include<stdio.h>
# include<string.h>
# include<algorithm>
using namespace std;
struct node{
node * next[10];
node(){
memset(next,0,sizeof(next));
}
};
int flag;
struct nod{
char s[10];
int len ;
}a[100005];
bool cmp(nod a,nod b)
{
return a.len>b.len;
}
void insert(char *s,node * & root) //插入
{ flag=0; //用來標記是否有重複的
node *p=root;
int i,k;
for(i=0;s[i];i++)
{
k=s[i]-'0';
if(p->next[k]==NULL)
{
p->next[k]=new node();
flag=1;
}
p=p->next[k];
}
}
int main(){
int t;
scanf("%d",&t);
while(t--)
{
int n;
node *root=new node();
scanf("%d",&n);
flag=1;
for(int i=0;i<n;i++)
{
scanf("%s",a[i].s);
a[i].len=strlen(a[i].s);
}
sort(a,a+n,cmp); //將長度排序
for(int i=0;i<n;i++)
{
if(flag)
insert(a[i].s,root);
else break;
}
if(flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}
