HDOJ1671 字典树入门题+模板程序（指针实现）

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21253    Accepted Submission(s): 7184

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）

 

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代码如下：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <ctime>
#include <cstdio>
using namespace std;

const int maxn = 1e4+10;
struct node{
     int cnt;
     node *next[10];
};

node *root;

struct word{
      char phone[15];
      int len;
}a[maxn];

bool cmp(word p ,word q){
   return p.len<q.len;
}

void create(char *str){
     int len=strlen(str);
     node *p=root,*q;
     for (int i=0;i<len;i++){
        int in=str[i]-'0';
        if (p->next[in]==NULL){
            q=(node *)malloc(sizeof(node));
            q->cnt=1;
            for (int k=0;k<10;k++) q->next[k]=NULL;

            p->next[in]=q;
            p=p->next[in];
        }
        else {
            p->next[in]->cnt++;
            p=p->next[in];
        }
     }
}

bool query(char *str){
    int len=strlen(str);
    node *p=root, *q;
    for (int i=0;i<len ;i++){
        int in=str[i]-'0';
        p=p->next[in];
        if (p==NULL) return 0;
    }
    return p->cnt;
}

void freetree(node *str){  //释放内存，不然会爆掉
    if (str==NULL) return;
    for (int i=0;i<10;i++){
        if (str->next[i]!=NULL) freetree(str->next[i]);
    }
    free(str);
    return;
}

int main(){
    std::ios::sync_with_stdio(false); //提升cin输入效率
    int t,n;
    cin >> t;
    while(t--){
        cin >> n;
        for (int i=0;i<n;i++) {
            cin>>a[i].phone;
            a[i].len=strlen(a[i].phone);
        }
        sort(a,a+n,cmp); //比较的是word类型，所以要自定义一个cmp函数

        root=(node *)malloc(sizeof(node)); //动态申请内存，所以最后要释放
        for (int i=0;i<10;i++) root->next[i]=NULL; //root不是真的字符串，所以root的cnt设置为零
        create(a[n-1].phone); //从长度最大的字符串开始插入
        bool flag=1;

        for (int i=n-2;i>=0;i--)
            if (query(a[i].phone)==0){
                 create(a[i].phone);
            }
            else{
                flag=0;
                break;
            }

        if (flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
        freetree(root); //释放内存
    }
    return 0;
}