HDU1671(Phone List)

Phone List

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

思路

题目大概意思就是判断一个号码是不是其他号码的前缀。直接上字典树 或者 map。字典树做法就是给前缀数目统计，并且给号码进行标记。如果一个前缀次数 >= 2次同时这个前缀是一个号码，那么输出NO，否则输出YES。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <map>
using namespace std;
const int maxn = 100005;
int trie[maxn][10];
int num[maxn];
bool f[maxn];
int cnt;
void insert_s(string s)
{
	int n = s.length();
	int root = 0;
	for(int i = 0;i < n;i++){
		int k = s[i] - '0';
		if(!trie[root][k]){
			trie[root][k] = ++cnt;
		}
		root = trie[root][k];
		num[root]++;
	}
	f[root] = true;
}
int main()
{
	int t,n;
	cin >> t;
	while(t--){
		cin >> n;
		string s;
		cnt = 0;
		memset(trie,0,sizeof(trie));
		memset(f,false,sizeof(f));
		memset(num,0,sizeof(num));
		for(int i = 0;i < n;i++){
			cin >> s;
			insert_s(s);
		}
		bool flag = true;
		for(int i = 1;i <= cnt;i++){
			if(num[i] >= 2 && f[i]){			//前缀数目 >= 2，同时这个前缀是一个号码。
				flag = false;
			}
		}
		if(flag){
			cout << "YES" << endl;
		}
		else{
			cout << "NO" << endl;
		}
	}
	return 0;
} 

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