nyoj163phone list 字典树

Phone List

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

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.

输入

The first line of input gives a single integer, 1 ≤ t ≤ 10, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 ≤ n ≤ 100000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.

输出

For each test case, output "YES" if the list is consistent, or "NO" otherwise.

样例输入

2391197625999911254265113123401234401234598346

样例输出

NO
YES

此处附上讲解比较好的字典树的模版代码http://www.cnblogs.com/tanky_woo/archive/2010/09/24/1833717.html

 #include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX 11typedef struct node{    int flag;//可记录字典树中某一串是否到结尾    struct node *next[10];}Lnode;//初始化Lnode *creat_list(){    Lnode *head;    int i;    head = (Lnode *)malloc(sizeof(Lnode));    if(head == NULL)    {        printf("no memory available\n");        exit(1);    }    for(i = 0; i < 10; i++)    {        head->next[i] = NULL;    }    head->flag = 0;    return head;}//构建插入字典树int insert_list(Lnode *root, char *s){    int i, flag = 0;    Lnode *phead = root;    for(i = 0; s[i] != 0; i++)    {        if(phead->next[s[i] - '0'] == NULL)        {            phead->next[s[i] - '0'] = creat_list();            flag = 1;        }        phead = phead->next[s[i] - '0'];        if(phead->flag == 1)            return 0;//说明该字符串还么有创建完就发现字典树以到末尾，说明此字典树中有该字符串的前缀，输出No    }    phead->flag = 1;//一个单词结尾标志赋值为1    if(flag == 0)        return 0;//说明没有重新构建结点，说明该字符串是此字典树中某一字符串的前缀，输出No    return 1;//说明相互无前缀}//释放内存void delete_list(Lnode *head){    int i;    if(head != NULL)    {        for(i = 0; i < 10; i++)        {            delete_list(head->next[i]);        }        free(head);        head = NULL;    }}int main(){    int n, num, flag;    scanf("%d", &n);    char s[MAX];    Lnode *root;    while(n--)    {        flag = 1;        scanf("%d", &num);        root = creat_list();        while(num--)        {            getchar();            scanf("%s", s);            if(flag == 1)                flag = insert_list(root, s);        }        delete_list(root);        if(flag == 0)            printf("NO\n");        else            printf("YES\n");    }    return 0;}