山东第一届省赛1001 Phone Number（字典树）

Phone Number

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

We know that if a phone number A is another phone number B’s prefix, B is not able to be called. For an example, A is 123 while B is 12345, after pressing 123, we call A, and not able to call B.
Given N phone numbers, your task is to find whether there exits two numbers A and B that A is B’s prefix.
 

输入

 The input consists of several test cases.
 The first line of input in each test case contains one integer N (0<N<1001), represent the number of phone numbers.
 The next line contains N integers, describing the phone numbers.
 The last case is followed by a line containing one zero.

输出

 For each test case, if there exits a phone number that cannot be called, print “NO”, otherwise print “YES” instead.

示例输入

2
012
012345
2
12
012345
0

示例输出

NO
YES

提示

 

来源

 2010年山东省第一届ACM大学生程序设计竞赛

题意：给出几个字符串问是否会有前缀产生，直接字典树搞定

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int Max =  + ;
 struct Node
 {
     int value;
     int cnt;
     Node * Next[];
 };
 Node * root;
 char str[Max];
 bool build(char * s)
 {
     Node * p = root, *temp;
     int len = strlen(s);
     for (int i = ; i < len; i++)
     {
         int id = str[i] - '';
         if (p->Next[id] == NULL)
         {
             temp = new Node;
             temp->value = id;
             temp->cnt = ;
             for (int i = ; i < ; i++)
                 temp->Next[i] = NULL;
             p->Next[id] = temp;
         }
         p = p->Next[id];
         if (p->cnt)
             return false;
     }
     return true;
 }
 void destroy(Node * temp)
 {
     if (temp == NULL)
         return;
     for (int i = ; i < ; i++)
         destroy(temp->Next[i]);
     delete temp;
 }
 int main()
 {
     int n;
     while (scanf("%d", &n) != EOF && n)
     {
         root = new Node;
         for (int i = ; i < ; i++)
             root->Next[i] = NULL;
         bool flag = true;
         while (n--)
         {
             scanf("%s", str);
             if (!flag)
                 continue;
             if (!build(str))
                 flag = false;
         }
         if (flag)
             printf("YES\n");
         else
             printf("NO\n");
         destroy(root);
     }
     return ;
 }