Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11669 Accepted Submission(s): 3970
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：给出一些数字，判断这些数字中是否有数字是其他数字的前缀。
分析：字典树
1 #include<cstring>
2 #include<cstdio>
3 #include<iostream>
4 #include<string>
5
6 using namespace std;
7 #define maxn 1000900
8
9 int dic[maxn][15],val[maxn],flag,cnt;
10
11 void inst(string s)
12 {
13 int u=0,v,len;
14 len=s.length();
15 for(int i=0;i<len;i++)
16 {
17 v=s[i]-‘0‘;
18 if(!dic[u][v])
19 {
20 dic[u][v]=++cnt;// 构造链表
21 memset(dic[cnt],0,sizeof(dic[cnt]));//优化
22 }
23 u=dic[u][v]; //u节点与v节点相连 dic的值指向下一个节点
24 if(i==len-1) val[u]++;// 标记该字符串结束位置
25 }
26 }
27
28 void get(string s)
29 {
30 int u=0,v,len;
31 len=s.length();
32 for(int i=0;i<len-1;i++)
33 {
34 v=s[i]-‘0‘;
35 if(val[dic[u][v]]) //如果该字符串结束之前发现有其他字符串结束点 说明产生前缀
36 {
37 flag=0;
38 return ;
39 }
40 u=dic[u][v];
41 }
42 }
43
44 int main()
45 {
46 int T;
47 scanf("%d",&T);
48 while(T--)
49 {
50 int n;
51 string s[100000];
52 scanf("%d",&n);
53 memset(dic[0],0,sizeof(dic[0]));
54 memset(val,0,sizeof(val));
55 cnt=0;
56 flag=1;
57 for(int i=0;i<n;i++)
58 {
59 cin>>s[i];
60 if(flag) inst(s[i]); // 这里以及下面最好都优化一下 不然会T
61 if(flag) get(s[i]);
62 }
63 for(int i=0;i<n&&flag;i++)
64 if(flag) get(s[i]);
65 else break;
66 if(flag) printf("YES\n");
67 else printf("NO\n");
68 }
69 }