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Phone List(字典树)

2024-07-19 23:03:23   220人阅读

Phone ListCrawling in process...Crawling failedTime Limit:1000MS    Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 
题意：看是否有相同的前缀，如果有输出NO，没有输出YES
ps：注意要释放内存。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
using namespace std;
struct node
{
    int flag;
    struct node *next[10];
}tree;
struct node *root;
void create_tree(char *str)
{
    int i,j;
    int len=strlen(str);
    struct node*p=root, *q;
    for(i=0;i<len;i++)
    {
        int id=str[i]-'0';
        if(p->next[id]==NULL)
        {
            q=(struct node*)malloc(sizeof(struct node));
            q->flag=1;//初始的flag==1；
            for(j=0;j<10;j++)
                q->next[j]=NULL;
            p->next[id]=q;
            p=p->next[id];
        }
        else
        {
            p->next[id]->flag++;
            p=p->next[id];
        }
    }
    p->flag=-1;//若为结尾，则将v改成-1表示

}

int find_tree(char *str)
{
    int i;
    int len=strlen(str);
    struct node *p=root;
    for(i=0;i<len;i++)
    {
        int id=str[i]-'0';
        p=p->next[id];
        if(p==NULL)//若为空集，表示不存以此为前缀的串
          return 0;
        if(p->flag==-1)//字符集中已有串是此串的前缀
            return -1;
    }
    return -1;//此串是字符集中某串的前缀
}

int freedom(struct node * T)//释放空间
{
    int i;
    if(T==NULL)
        return 0;
    for(i=0;i<10;i++)
    {
        if(T->next[i]!=NULL)
            freedom(T->next[i]);
    }
    free(T);
    return 0;
}

int main()
{
    char str[20];
    int n,m,i,j;
    int flag;
    scanf("%d", &n);
    while(n--)
    {
        flag=0;
        root=(struct node *)malloc(sizeof(struct node));
        for(i=0;i<10;i++)
            root->next[i]=NULL;
            root->flag=0;
        scanf("%d", &m);
        while(m--)
        {
            scanf("%s",str);
              if(find_tree(str)==-1)
                flag=1;
                if(flag)
                continue;
              create_tree(str);

        }
        if(flag)
            printf("NO\n");
        else
            printf("YES\n");
        freedom(root);
    }
    return 0;
}



Phone List(字典树)


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