sicily 1426 PhoneList 使用动态分配是在是太耗时了

1426. Phone List

  

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012345

    

    

Time Limit: 1sec    Memory Limit:64MB

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:
• 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 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

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2391197625999911254265113123401234401234598346

Sample Output

NOYES

下面这个程序用时0.24，关键是动态分配耗时比较多，且使用两个指针，空间也不理想，解决办法为申请一块数组，

TreeNode data[1000000],

每次分配时不用new,而直接映射到数组就好，这样用时为0.07秒（已测试），大大提高（这种方法对于这道题的确很好，

不过数组开多大，你自己试罗，我60009就过

样例如此而已）

#include<iostream>#include<string.h>#include<stdlib.h>using namespace std;class TreeNode;typedef TreeNode *Tree;typedef Tree Position;/*  Implementation  */class TreeNode{public:Tree DownTree;Tree RightTree;char digt;TreeNode(){DownTree = RightTree = NULL;}}; /* *用法:int (char* PhoneNum, Tree Root) *以Root为根往下初始化一棵树，数的每个节点以PhoneNum[0]给digt赋值 *若根不为空，则在根的右树初始化 */Tree& ini(char PhoneNum[],Tree& Root){if(Root != NULL)//根不为空return ini(PhoneNum, Root->RightTree);if(PhoneNum[0] == '/0')//树建立完毕，返回NULLreturn Root;//否则，继续建立下一级树DownTreeRoot =new TreeNode;Root->digt = PhoneNum[0];ini(&PhoneNum[1], Root->DownTree);}/* *使用：Find(char item, Tree Cur) *在Cur所在的这一级树进行查找item *若找到，返回所在位置的指针，否则，返回NULL */inline Position& FindInSameLevel(char& item, Tree& Cur){if(Cur  == NULL)//Cur为空，该级未初始化,返回空指针return Cur;do{//否则，沿着DownTree查找if(Cur->digt == item)return Cur;else    Cur = Cur->RightTree;}while(Cur != NULL);return Cur;}/* *用法：isPrefix(char PhoneNum[], Tree Root) *判断以Root为根的数是否存在前缀现象 *若有，返回True *否则，返回false */bool isPrefix(char PhoneNum[], const Tree& Root){Tree Cur = Root;bool flag = false;//flag is true if there's prefixfor(int i=0; i < strlen(PhoneNum); i++){Tree tmpCur = Cur;//记录Cur，若Cur被定为为NULL，则对其重定位if(FindInSameLevel(PhoneNum[i], Cur)){flag = true;//找到一个对应的就可以把flag设置为trueif(FindInSameLevel(PhoneNum[i], Cur)->DownTree != NULL)//找到，且该点不为末端{Cur = FindInSameLevel(PhoneNum[i], Cur)->DownTree;continue;}elsereturn true;//该点为末端，树中数据为PhoneNum前缀}else//找不到该点，则初始化一棵树{Cur= tmpCur ;flag = false;ini(&PhoneNum[i], Cur->RightTree);break;//初始化完了则继续处理下一个号码，不判断flag}}if(flag)return true;elsereturn false;}int main(){int TestCase;cin>>TestCase;while( TestCase-- ){int NumofPhoneNum;cin>>NumofPhoneNum;Tree Root; Root =NULL;bool flag;  flag= true;//true 如果没前缀for(int i=0; i< NumofPhoneNum; i++){char PhoneNum[10];cin>>PhoneNum;if(i == 0){ini(PhoneNum, Root);//初始化第一个号码}elseif(isPrefix(PhoneNum, Root)) //判断后面输入的号码是否构成前缀{flag = false;char rubbish[10];for(int j=0; j < NumofPhoneNum - i-1; j++)//构成了前缀，继续接收完剩余的PhoneNumcin>>rubbish;cout<<"NO"<<endl;break;}}if(flag)//不构成，输出YEScout<<"YES"<<endl;}return 0;}