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Phone List hdu1671 trie

2019-11-08 01:39:40
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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.

Source


2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up(3)

Solution


trie求前綴例題。如果一個電話的最后一個節點仍然有后繼節點說明它是某另外一個電話的前綴嗯

Code


#include <iostream>#include <string.h>#include <string>#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)#define fill(x, t) memset(x, t, sizeof(x))#define N 100001#define L 58using namespace std;int map[N][L], p[N], cnt;bool flag;string ans;inline void insert(int now, string s, int dep){ if (dep == s.length()){ p[now] += 1; return; } int tar = s[dep]; if (!map[now][tar]){ map[now][tar] = ++ cnt; } insert(map[now][tar], s, dep + 1);}inline void query(int now, string s){ if (flag){ return; } rep(tar, '0', '9'){ if (map[now][tar]){ if (p[now]){ ans = "NO"; flag = true; return; } string tmp = s; tmp += tar; query(map[now][tar], tmp); } }}int main(void){ ios::sync_with_stdio(false); int T; cin >> T; while (T --){ cnt = 0; ans = "YES"; flag = false; fill(map, 0); fill(p, 0); int n; cin >> n; rep(i, 1, n){ string s; cin >> s; insert(0, s, 0); } query(0, ""); cout << ans << endl; } return 0;}
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