POJ-3630 Phone List（字典樹）

P - Phone List

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit 

Status

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.

For each test case, output “YES” if the list is consistent, or “NO” otherwise.

2
3
911
97625999
91125426
5
113
12340
123440
12345
98346      

NO
YES      

題意：給你n個電話号碼，判斷是否存在一個電話号碼是另一個号碼的字首，如果存在則輸出no，否則輸出yes

#include<stdio.h>
#include<string.h>
#include <iostream>  
#include <string>  
#include <algorithm>   
using namespace std;  
  
int ct;  
struct TireNode  
{  
       int count;  
       struct TireNode *branch[10];  
       TireNode()  
       {  
                 count = 0;  
                 for (int i = 0; i < 10; i++)  
                      branch[i] = NULL;   
       }   
}NodeNum[100000];//靜态數組儲存Tire樹結點！   
  
//Tire樹的插入操作！   
void TireInsert(TireNode *&root, string str)  
{  
     int i, len;  
     len = str.length();  
     TireNode *p = root;   
     for (i = 0; i < len; i++){  
         if (!p->branch[str[i]-'0']){  
             p->branch[str[i]-'0'] = &NodeNum[ct];  
             ct++;  
         }  
         p = p->branch[str[i]-'0'];  
         p->count++;   
     }   
     return ;   
}   
  
//Tire樹的查詢操作！   
bool TireSearch(TireNode *root, string str)  
{  
     if (root == NULL)  
         return 0;  
     int i, len;  
     len = str.length();  
     TireNode *p = root;  
     for (i = 0; i < len; i++){  
         p = p->branch[str[i]-'0'];   
         if (p->count == 1)  
             return 1;   
     }   
     return 0;   
}   
  
  
int main()  
{  
    int tc, i, n;  
    string phone[10010];   
    bool flag;   
    cin >> tc;  
    while (tc--){  
          cin >> n;  
          flag = true;   
          memset(NodeNum, NULL, sizeof(NodeNum));  
          TireNode *root = &NodeNum[0];  
          ct = 1;  
          for (i = 0; i < n; i++){  
              cin >> phone[i];  
              TireInsert(root, phone[i]);   
          }   
          sort(phone, phone+n);   
          for (i = 0; i < n; i++){  
              if (!TireSearch(root, phone[i])){  
                   flag = false;  
                   break;   
              }   
          }   
          if (flag)  
              cout << "YES" << endl;  
          else  
              cout << "NO" << endl;   
    }   
      
    system("pause");   
}