zoj 2876 Phone List(tire 树)

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Phone List

Time Limit: 5 Seconds      Memory Limit: 32768 KB

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 withn, the number of phone numbers, on a separate line, 1 <= n <= 10000.Then followsn 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

#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#define N 10using namespace std;struct trie{    struct trie *next[N];    bool isword;    int num;}*head;char temp[100];int flag=0;struct trie* hiti(){    struct trie* node = (struct trie *)malloc(sizeof(struct trie));    node->num = 0;    node->isword = false;    memset(node->next , 0 , sizeof(node->next));    return node;}void great_tree(char *a){    struct trie *p = head,*q = NULL;    int len = strlen(a);    for(int i=0;i<len;i++)    {        int id = a[i]-'0';        if(p->next[id]==NULL)        {            q = new (struct trie);            for(int j=0;j<N;j++)                q->next[j] = NULL;            q->isword = 0;            q->num = 0;            p->next[id] = q;        }        if(p->next[id]->isword) flag = 1;        if(i==len-1)        {            p->next[id]->isword = 1;        }        p = p->next[id];    }    for(int i=0;i<N;i++)    if(p->next[i]!=NULL) flag = 1;}void del(struct trie *head){    for(int i=0;i<N;i++)    {        if(head->next[i]!=NULL)            del(head->next[i]);    }    delete head;}int find_tree(char a[]){    struct trie *p = head;    int len = strlen(a);    for(int i=0;i<len;i++)    {        int id = a[i]-'0';     //   if(p->next[id]==NULL)      //      return -1;        p = p->next[id];    }    if(p->isword)        return p->num;}int main(){    int Case,n;    cin >> Case;    getchar();    while(Case--)    {        cin >> n ;        getchar();        flag = 0;        head = hiti();        for(int i=0;i<n;i++)        {            gets(temp);            great_tree(temp);        }        if(!flag)            cout << "YES" << endl;        else            cout << "NO"  << endl;        del(head);    }    return 0;}


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