HDU 1671 Phone List

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Phone List

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 296 Accepted Submission(s): 120

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

2391197625999911254265113123401234401234598346

Sample Output

NOYES

Source

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

题目大意：

给出一些电话号，要求不能出现某个电话号是别的电话号的前缀。

思路：

加上一个结束的标志就好了。

AC代码：

#include<iostream>#include<cstring>#include<queue>#include<cstdio>#include<cmath>#include<algorithm>#include<vector>using namespace std;const int MAXN = 10;struct Trie{    Trie *Next[MAXN];    int Flag;    Trie()    {        Flag = 0;        for(int i=0;i<10;i++)        {            Next[i]=NULL;        }    }}*root;int Insert(char *str){    int len = strlen(str);    Trie *p = root, *q;    for(int i=0; i<len; ++i)    {        int id = str[i] - '0';        if(p->Next[id] == NULL)        {            q = new Trie();            p->Next[id] = q;        }        else        {            if(p->Next[id]->Flag||str[i+1]=='\0')                return 1;        }        p = p->Next[id];    }    p->Flag=1;    return 0;}void Free(Trie* T)//删除{    for(int i=0; i<MAXN; ++i)    {        if(T->Next[i]) Free(T->Next[i]);    }    free(T);}int main(){    int T;    scanf("%d",&T);    int n;    char ch[30];    while(T--)    {        scanf("%d",&n);        int fa=0;        root=new Trie();        for(int i=0;i<n;i++)        {            scanf("%s",ch);            if(fa)                continue;            if(Insert(ch))            {                fa=1;            }        }        if(fa)            cout<<"NO"<<endl;        else            cout<<"YES"<<endl;        Free(root);    }}

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