| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 20894 | Accepted: 6532 |
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:
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：判断每组串是不是某个串的前缀。
sl:字典树轻松解决，沿途判断是不是经过单词节点，如果当前单词判断完了那么判断当前节点是不是有后继节点
ps:差点爆内存，最好用左儿子右兄弟。
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 const int MAX = 20;
6 const int A = 1000000;
7 char num[MAX];
8 int flag;
9 int ch[A][11];
10 struct Trie
11 {
12 int val[A]; int sz;
13 Trie(){sz=1; memset(ch[0],0,sizeof(ch[0])); }
14 int index(char a) {return a-‘0‘;}
15 void insert(char *s,int v)
16 {
17 int u=0,c; int n=strlen(s);
18 for(int i=0;i<n;i++)
19 {
20 c=index(s[i]);
21 if(!ch[u][c])
22 {
23 memset(ch[sz],0,sizeof(ch[sz]));
24 val[sz]=0;
25 ch[u][c]=sz++;
26 }
27 u=ch[u][c];
28 if(val[u]!=0) flag=1;
29 }
30 val[u]=v;
31 for(int i=0;i<10;i++) if(ch[u][i]) flag=1;
32 }
33 };
34 int main()
35 {
36 int cas,n;
37 scanf("%d",&cas);
38 while(cas--)
39 {
40 Trie trie;
41 scanf("%d",&n); flag=0;
42 for(int i=0;i<n;i++)
43 {
44 scanf("%s",num);
45 trie.insert(num,i+1);
46 }
47 if(flag) printf("NO\n");
48 else printf("YES\n");
49 }
50 }
51 /*
52 5 5
53 123456
54 6656
55 3123
56 13646
57 1564
58 */
原文：http://www.cnblogs.com/acvc/p/3583320.html
