POJ 3630 Phone List 【判断串是否是其他串的前缀】

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本文介绍了一个电话号码列表的一致性检查算法，确保任一号码不是其他号码的前缀，避免呼叫混淆。通过构建Trie树并遍历每个号码来判断其是否为其他号码的前缀。

Phone List

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28119		Accepted: 8412

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

题意：给定字符串集合，判断是否有某个串为别的串的前缀；

思路：构造出Trie，然后枚举每一个串判断这个串是否为其他串的前缀,枚举所有即可;

失误：1.数组开小了，少了一位，一直runtime error;

2.过了一下940ms 看宇神的思路一样他的就一百多秒，又是数组开大了，改小之后明显时间空间都小了，看来数组开的也要适度；

AC代码：

#include<cstdio>
#include<cstring>

struct tree{
	int num;
	int son[17];
}Node[110111];
char S[10111][14];
int sn=0;

void init()
{
	memset(Node,0,sizeof(Node)); sn=0;
}
void insert(char *str)
{
	int i=0,p=0;
	while(str[i]) {
		int pos=str[i]-'0';
		if(Node[p].son[pos]==0) 
		   Node[p].son[pos]=++sn;
		p=Node[p].son[pos]; ++i;
		++Node[p].num;
	}
} 	
bool Query(char *str)
{
	int i=0,p=0;
	while(str[i]) {
		int pos=str[i]-'0';
		p=Node[p].son[pos]; ++i;
		if(Node[p].num==1) 
		   return true;
	}
	return Node[p].num>1?false:true;
}
int main()
{
	int T,i,N;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&N); init();
		int i=0,cnt=0;
		for(i=1;i<=N;++i) 
		{
			scanf("%s",S[++cnt]); 
			insert(S[cnt]);
		}
		bool falg=false;
		for(i=1;i<=N;++i)
		{
			if(!Query(S[i])) {
				falg=true;  break;
			}
		}
		if(falg) printf("NO\n");
		else printf("YES\n");
	}
	return 0;
}