如何用字典树高效查询长尾词在电话簿中的出现？

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电话列表时间限制：3000/1000 MS（Java/其他）内存限制：32768/32768 K（Java/其他）总提交次数：15853接受提交次数：5326问题描述：给定一组电话号码，判断该列表是否在一致性方面满足条件。

Phone List

Time Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 15853Accepted Submission(s): 5326

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

Source

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

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代码：

#include<cstdio>
#include<cstring>
#define ms(a,x) memset(a,0,sizeof(a[0])*(x+5))
using namespace std;

const int maxn=1e4;
const int max_len=10;
int num,f[maxn*max_len][10];
bool v[maxn*max_len];

int main()
{
//freopen("1.in","r",stdin);

int i,j,t,n,len;bool flag;char s[max_len+5];
for(scanf("%d",&t);t;t--)
{
ms(f,num),ms(v,num),flag=num=0;
for(scanf("%d",&n);n && scanf("%s",s);n--)
{
if(flag)continue;
len=strlen(s);
for(i=0,j=0;!v[i] && j<len;i=f[i][s[j++]-'0'])
if(f[i][s[j]-'0']==0)f[i][s[j]-'0']=++num;
if(flag=v[i]++)continue;
for(j=0;j<10;j++)if(f[i][j]){flag=1;break;}
}
printf("%s\n",flag?"NO":"YES");
}
return 0;
}