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 911Alice 97 625 999Bob 91 12 54 26In 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 98346Sample Output
NO YES #include <iostream>#define MAX_N 120000using namespace std;int trieTree[MAX_N + 1][12]; //第10位是ID,第11位是countint countv = 0;bool insert(char phone[], int id){int curNode = 0, i, next;char c;for(i = 0; i < strlen(phone); i++) { c = phone[i]; trieTree[curNode][11]++; next = trieTree[curNode][c - '0'];if(next == 0) trieTree[curNode][c- '0'] = next = ++countv; curNode = next;if(trieTree[curNode][10] != 0)return false; }if(trieTree[curNode][11] > 0)return false; trieTree[curNode][10] = id;return true;}int main(){ int caseNum, c, i, pNum;char temp[10]; scanf("%d", &caseNum);for(c = 0; c < caseNum; c++) { memset(trieTree, 0, sizeof(trieTree)); countv = 0; scanf("%d", &pNum);bool can = true;for(i = 0; i < pNum; i++) { scanf("%s", temp);if(can)if(!insert(temp, i + 1)) can = false; }if(can) printf("YES\n");else printf("NO\n"); }return 0;}转载于:https://www.cnblogs.com/w0w0/archive/2011/11/23/2259854.html
