比较惊险的一次字典树
继上次动态分配内存字典树，这次尝试使用静态数组存放字典树：每个trie[i][j]保存其子树的节点序号，使用全局变量更新，权值以及其余值则另外存储
题意：判断给出的几个数字串中是否有一个是另一个的前缀
使用字典树比较时需要处理 123 123， 123 123456， 123 12,1230 1234等情况，细节较多，不过还是可以通过添加辅助数组以及标志变量处理的。
注：process描述可能有点差错，只供参考（反正我当时是看着写的）
代码：
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 #include <set> 6 #include <queue> 7 using namespace std; 8 #define maxnode 12 9 #define maxn 10005 10 11 /* 12 process: 13 string s; 14 flag 15 along the tree 16 s[i] already exist? 17 i == s_size? return 0; 18 else go to it's sub tree 19 No s[i] ? 20 i != 0? 21 flag & no hasChild? return 0; 22 create node, refresh flag 23 i == 0? 24 create node, refresh flag 25 */ 26 int trie[maxn * 10][maxnode];//0 as root 27 bool hasChild[maxn * 10]; 28 int totindex = 1; 29 char st[maxn][15]; 30 31 void init(){ 32 memset(trie, 0, sizeof(trie)); 33 memset(st, 0, sizeof(st)); 34 memset(hasChild, 0, sizeof(hasChild)); 35 totindex = 1; 36 } 37 38 int insert_find(char tms[]){ 39 bool flag = true; 40 int len = strlen(tms), i = 0, j = 0; 41 for(; i < len; i++){ 42 if(trie[j][tms[i] - '0']){ 43 if(i == len - 1) 44 return 0; 45 j = trie[j][tms[i] - '0']; 46 } 47 else{ 48 if(i != 0){ 49 if(flag && !hasChild[j]){ 50 return 0; 51 } 52 else{ 53 hasChild[j] = true; 54 flag = false; 55 trie[j][tms[i] - '0'] = totindex++; 56 j = trie[j][tms[i] - '0']; 57 } 58 } 59 else{ 60 hasChild[j] = true; 61 flag = false; 62 trie[j][tms[i] - '0'] = totindex++; 63 j = trie[j][tms[i] - '0']; 64 } 65 } 66 } 67 return 1; 68 } 69 70 int main(){ 71 int t; 72 scanf("%d", &t); 73 while(t--){ 74 int n; 75 init(); 76 memset(st, 0, sizeof st); 77 scanf("%d", &n); 78 for(int i = 0; i < n; i++){ 79 scanf("%s", st[i]); 80 } 81 82 bool flag = true; 83 for(int i = 0; i < n; i++){ 84 int tmp = insert_find(st[i]); 85 if(!tmp){ 86 flag = false; 87 break; 88 } 89 } 90 91 if(flag){ 92 printf("YES "); 93 } 94 else{ 95 printf("NO "); 96 } 97 } 98 }
题目：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 29377 | Accepted: 8764 |
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:
- Emergency 911
- Alice 97 625 999
- 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