HDU 1671 Phone List （字典树）

最新推荐文章于 2020-05-09 10:07:54 发布

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最新推荐文章于 2020-05-09 10:07:54 发布

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分类专栏： ACM_字典树 ACM_链表文章标签：字典树

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本文介绍了一种使用字典树解决电话号码簿中号码是否互为前缀的问题的方法。通过构建字典树来检查每个号码是否是另一个号码的前缀，并提供了一个具体的实现示例。

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Phone List

题目链接

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21088 Accepted Submission(s): 7113

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）

题意：判断输入的电话号码中是否有号码是其他号码的前缀，如果有输出NO，否则输出 YES 。经典的字典树。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
    int num;
    node *next[12];
    node()//初始化
    {
        num=0;
        for(int i=0;i<12;i++)
            next[i]=NULL;
    }
};
int state;
node *root;
void Buildtree(char *a)
{
    node *p=root;
    int len=strlen(a);
    int t=0,f=0;
    for(int i=0;i<len;i++)
    {

        if(p->next[a[i]-'0']!=NULL) t++;//顺这字典树，统计号码相同的个数
        if(p->num==1) f=1;//这个是用来判断前面输入的是否有现在输入的号码的前缀的
        if(p->next[a[i]-'0']==NULL)
            p->next[a[i]-'0']=new node();//如果为空申请新的空间，t的值将不会为len也就是说这个号码不是前面所有号码的前缀
        p=p->next[a[i]-'0'];
    }
    p->num=1;//当一个号码输入完将那个节点的num赋值为1
    if(t==len||f==1)
        state=1;
}
void delett(node *p)//清空字典树
{
    for(int i=0;i<12;i++)
        if(p->next[i])
            delett(p->next[i]);
    delete(p);
}
char a[12];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        state=0;
        root=new node();
        while(n--)
        {
            scanf("%s",a);
            if(state==1)//已经有一个号码是别的号码的前缀了，后面就没必要接着建字典树了
                continue;
            Buildtree(a);
        }
        if(state==1)
            printf("NO\n");
        else
            printf("YES\n");
        delett(root);//将字典树清空，不然后超内存的
    }
    return 0;
}