Phone List
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18421 Accepted Submission(s): 6207
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
题意：
一次给出n个电话号码，如果刚给出的一个电话号码的前几位包含前面给出的某一个号码则这个号码不能拨打（输入那前几位时就会马上打出去了）。
代码：
1 //字典树的权值作为标记点，每输入一个号码，查找他每一位的val值有没有等于1的，如果没有就把他放入字典树他的最后一
2 //位的val值置为1，如果有说明与前面的某一个冲突。
3 #include<iostream>
4 #include<cstdio>
5 #include<cstring>
6 using namespace std;
7 const int MAX=1000000;
8 const int CON=10;
9 int nod[MAX][CON],val[MAX];
10 int sz;
11 void init()
12 {
13 sz=1;
14 memset(nod[0],0,sizeof(nod[0]));
15 val[0]=0;
16 }
17 void insert(char s[])
18 {
19 int len=strlen(s);
20 int rt=0;
21 for(int i=0;i<len;i++)
22 {
23 int id=s[i]-‘0‘;
24 if(nod[rt][id]==0)
25 {
26 memset(nod[sz],0,sizeof(nod[sz]));
27 nod[rt][id]=sz;
28 val[sz++]=0;
29 }
30 rt=nod[rt][id];
31 }
32 val[rt]=1;
33 }
34 int search(char s[])
35 {
36 int len=strlen(s);
37 int rt=0,cnt=0;
38 for(int i=0;i<len;i++)
39 {
40 int id=s[i]-‘0‘;
41 if(nod[rt][id]==0)
42 return 0;
43 rt=nod[rt][id];
44 if(val[rt]==1)
45 return 1;
46 }
47 }
48 int main()
49 {
50 char ch[20];
51 int t,n,flag;
52 scanf("%d",&t);
53 while(t--)
54 {
55 init();
56 scanf("%d",&n);
57 flag=0;
58 for(int i=1;i<=n;i++)
59 {
60 scanf("%s",ch);
61 if(flag) continue;
62 flag=search(ch);
63 insert(ch);
64 }
65 if(flag) printf("NO\n");
66 else printf("YES\n");
67 }
68 return 0;
69 }