【poj3630】Phone List 【Nordic 2007】

本文链接：https://blog.youkuaiyun.com/TLECOCE/article/details/76723988

给定一个电话号码列表，判断是否所有号码之间没有前缀关系。例如，如果911是紧急号码，而9112是另一个号码的前缀，那么这个列表就不一致。题目要求处理包含最多10000个电话号码的输入，并判断其是否一致。输出"YES"表示一致，"NO"表示不一致。

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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

Sample Output

NO
YES

这道题是一道tried树的模板题，下面是程序：

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
struct node{
	int x,next[10];
}m[100005];
struct num{
	int l;
	char a[15];
	bool operator <(num p)const {
		return l>p.l;
	}
}x[10005];
int k;
void add(int u,int w){
	m[u].next[w]=++k;
}
int dfs(int root,int h,num &p){
	int &q=m[root].next[p.a[h]-'0'];
	if(h==p.l-1){
		if(q==-1){
			add(root,p.a[h]-'0');
			return -1;
		}
		else{
			return q;
		}
	}
	if(q!=-1){
		return dfs(q,h+1,p);
	}
	add(root,p.a[h]-'0');
	dfs(q,h+1,p);
	return -1;
}
bool work(){
	k=0;
	memset(m,-1,sizeof(m));
	int n,i;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%s",x[i].a);
		x[i].l=strlen(x[i].a);
	}
	sort(x+1,x+n+1);
	for(i=1;i<=n;i++){
		if(dfs(0,0,x[i])!=-1){
			return 0;
		}
	}
	return 1;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		printf("%s\n",work()?"YES":"NO");
	}
	return 0;
}

注：要注意常数。