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HDU 1617 Phone List (排序||字典树)

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Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10720    Accepted Submission(s): 3701


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
 

 

Recommend
lcy
 

 

字符串排序再扫描一遍即可

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 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<stdlib.h>
 6 #include<algorithm>
 7 using namespace std;
 8 const int MAXN=10000+5;
 9 int T,n;
10 struct node
11 {
12     char str[10+5];
13 }a[MAXN];
14 
15 int cmp(const node A,const node B)
16 {
17     return strcmp(A.str,B.str)<0;
18 }
19 
20 int judge(char a[],char b[])
21 {
22     int len=strlen(a);
23     for(int i=0;i<len;i++)
24         if(a[i]!=b[i])
25             return 0;
26     return 1;
27 }
28 int main()
29 {
30     //freopen("in.txt","r",stdin);
31     scanf("%d",&T);
32     while(T--)
33     {
34         scanf("%d",&n);
35         for(int i=0;i<n;i++)
36             scanf("%s",a[i].str);
37 
38         sort(a,a+n,cmp);//利用结构体对字符串排序
39 
40         int ok=0;
41         for(int i=0;i<n-1;i++)
42         {
43             if(judge(a[i].str,a[i+1].str))
44             {
45                 ok=1;
46                 break;
47             }
48         }
49         
50         if(ok)
51             printf("NO\n");
52         else
53             printf("YES\n");
54     }
55     return 0;
56 }
View Code

 

HDU 1617 Phone List (排序||字典树)

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原文地址:http://www.cnblogs.com/clliff/p/3965269.html

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