poj3630 Phone List

Phone List

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 26137 Accepted: 7896

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

Source

Nordic 2007

Trie树模板题

注意这道题的字符串不是有序的，所以要全部插入后再判断。

据说快排也可以过...

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>#define F(i,j,n) for(int i=j;i<=n;i++)#define D(i,j,n) for(int i=j;i>=n;i--)#define ll long long#define pa pair<int,int>#define maxn 100005using namespace std;int tot=0,t,n;bool flag;struct trie_type{int tag,next[10];}tree[maxn];char s[10005][11];inline void insert(char *ch){int p=0,l=strlen(ch);F(i,0,l-1){int tmp=ch[i]-'0';if (!tree[p].next[tmp]) tree[p].next[tmp]=++tot;p=tree[p].next[tmp];}tree[p].tag++;}inline void judge(char *ch){if (!flag) return;int p=0,l=strlen(ch);F(i,0,l-2){int tmp=ch[i]-'0';p=tree[p].next[tmp];if (tree[p].tag){flag=false;return;}}}int main(){scanf("%d",&t);while (t--){F(i,0,tot){F(j,0,9) tree[i].next[j]=0;tree[i].tag=0;}tot=0;scanf("%d",&n);F(i,1,n){scanf("%s",s[i]);insert(s[i]);}flag=true;F(i,1,n) judge(s[i]);if (flag) puts("YES");else puts("NO");}}