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集训队专题(1)1003 Phone List

鱼板番茄 2022-08-30 阅读 92


Phone List

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 59   Accepted Submission(s) : 22

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

 


Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up(3)

 

       题目大意:给你几组电话号码,问你存不存在一个电话号码是另一个电话号码的前缀,如果有,则输出“NO”,否则输出“YES”,如样例1中911为91125426的前缀,所以输出“NO”。

       在这里小编用数组模拟去实现,当然也可以用指针去实现(小编没试过,不知道会不会爆内存~)。

       这里判断的条件还是有点坑,首先样例具有迷惑性,样例的前缀词出现在前面,所以很多人的判断条件都是只判断了最新建立的节点在之前建立的节点里是否存在前缀,但往往忽略了有可能前缀词在后面出现(别问小编怎么知道的,因为小编也是这样莫名其妙的WA了很多次TT)。判断新建立的节点是否是已建立的节点的前缀,只需要判断是否有新节点的出现即可。


#include <string>
#include <cstring>
#include <cstdio>
#include <iostream>
using namespace std;
struct node
{
bool end;//判断是否为号码的最后一位
int a[10];
};
bool x;
int ptr;
node list[100000];
void Init()
{
x = true;
ptr = 0;
for(int i=0; i<100000; i++)
{
list[i].end = false;
for(int j=0; j<10; j++)
list[i].a[j] = -1;
}
}
void Insert(char *s)
{
bool z=false;
bool y=false;
int now=0;
int len=strlen(s);
for(int i=0; i<len; i++)
{
if(list[now].a[s[i]-'0'] == -1)
{
z = true;
list[now].a[s[i]-'0'] = ++ptr;
now = ptr;
}
else
{
now = list[now].a[s[i]-'0'];
if(list[now].end) y = true;
}
}
list[now].end = true;
x=(!y)&&z;
}
int main()
{
int t,n;
char s[11];
scanf("%d",&t);
while(t--)
{
Init();
scanf("%d",&n);
while(n--)
{
scanf("%s",s);
if(x) Insert(s);
}
if(x) printf("YES\n");
else printf("NO\n");
}
return 0;
}



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