POJ-3630 Phone List

Phone List

Time Limit: 1000MS Memory Limit: 65536K   

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

2391197625999911254265113123401234401234598346

Sample Output

NOYES

————————————————————耐心的分割线————————————————————

思路：这个……字典树找前缀是吧……参见POJ-1056 IMMEDIATE DECODABILITY。这个思路是一样的。样式不同而已。

代码如下：

#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>using namespace std;struct TrieNode {TrieNode* next[10];bool word;}node[100010];char dic[10010][15];int nxt, flag;int cmp(const void* a, const void* b) {return strcmp((char*)a, (char*)b);}TrieNode *NewNode() {memset(&node[nxt], 0, sizeof(TrieNode));return &node[nxt++];}void Insert(TrieNode* rt, char* s) {int len = strlen(s);for(int i = 0; i < len; i++) {int c = s[i] - '0';if(!rt->next[c])rt->next[c] = NewNode();rt = rt->next[c];if(rt->word) {//一旦插入过程中遇见前缀，重复了，breakflag = 0;break;}}rt->word = true;}int main() {int cas;scanf("%d", &cas);while(cas--) {int m;scanf("%d", &m);nxt = 0; flag = 1;TrieNode* root = NewNode();//初始化字典树for(int i = 0; i < m; i++)scanf("%s", dic[i]);qsort(dic, m, sizeof(dic[0]), cmp);//先按字典序排个序,否则插入一个单词后再插入它的前缀，没反应。for(int i = 0; i < m; i++)Insert(root, dic[i]);printf("%s\n", flag ? "YES" : "NO");}return 0;}