Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9575    Accepted Submission(s): 3253

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  

之 所以想写此题的题解，就是想给自己一个警醒，动态内存，有分配就应该要有释放。

解析见代码：

#include <iostream>
#include <cstdio>
using namespace std;

const int maxx = 10;
char str[10000][10];

struct node
{
    int count; ///记录该节点被访问的次数
    node *next[maxx];
    node( ) ///使用New时自动调用，对NODE进行初始化
    {
        count = 1;
        for(int i = 0;i < maxx; i++)
        {
            next[i] = NULL;
        }
    }
};

void insert(node *&root,char *str)
{
    node *l=root;
    int i=0,branch;
    while(str[i]!='\0')
    {
        branch=str[i]-'0';
        if(l->next[branch])
        {
            l->next[branch]->count++;
        }
        else
        {
            l->next[branch]=new node; ///动态分配内存
        }
        l=l->next[branch];
        i++;
    }
}

int search(node *&root,char *str)
{
    node *l=root;
    int i=0,branch,sum;
    while(str[i]!='\0')
    {
        branch=str[i]-'0'; ///把字符型数字转换为整型数字
        if(l->next[branch])
        {
            l=l->next[branch];
            sum=l->count;
        }
        else return 0;
        i++;
    }
    return sum;
}

void del(node *p)  ///释放内存的函数
{
    int i=0;
    for(;i < 10;i ++)
    {
        if(p->next[i]!=NULL)  ///假如下一个节点不是NULL则继续寻找下一个节点的空节点
        {
            del(p->next[i]);
        }
    }
    delete p;
}

int main( )
{
    int n,t;
    cin>>t;
    while(t--)
    {
        node *root= new node; ///分配内存
        cin>>n;
        getchar( );
        for(int o=1;o <= n; o++)
        {
            gets(str[o]);
            insert(root,str[o]);
        }
        int judge = 0;
        for(int i=1; i<= n; i++)
        {
            if(search(root,str[i])>1)  ///一旦出现有相同前缀的字符串，则不必继续寻找
            {
                judge=1;
                break;
            }
        }
        if(judge)
        {
            cout<<"NO"<<endl;
        }
        else
        {
            cout<<"YES"<<endl;
        }
        del(root);  ///释放内存
    }
    return 0;
}