HDU_1671_Phone List（字典树）

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11885    Accepted Submission(s): 4040

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
   
   
   
   

题意：判断给定号码中是否存在某个号码是其他某号码的前缀，存在输出“NO”，否则输出“YES”。

解析：利用字典树处理。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1671

代码清单：

#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int MAX = 11;
struct trie{
    int point;
    trie *next[MAX];
};

trie *root;
int t,n,flag;
char s[MAX];

void Trie(char *s){
    trie *p=root, *q;
    int len=strlen(s), pos;
    for(int i=0;i<len;i++){
        pos=s[i]-'0';
        if(p->next[pos]==NULL){
            q=new trie();
            q->point=0;
            for(int j=0;j<MAX;j++)
                q->next[j]=NULL;
            p->next[pos]=q;
            p=p->next[pos];
        }
        else{
            if(p->next[pos]->point){ //前面某个号码是当前的前缀
                flag=1;
                return ;
            }
            p=p->next[pos];
        }
    }
    p->point=1;
    for(int i=0;i<MAX;i++){
        if(p->next[i]){   //当前号码是前面某个号码的前缀
            flag=1;
            return ;
        }
    }
}

void delTrie(trie *Root){  //释放内存空间
    for(int i=0;i<MAX;i++){
        if(Root->next[i]!=NULL)
            delTrie(Root->next[i]);
    }
    free(Root);
}

int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        root=new trie();
        for(int i=0;i<MAX;i++)
            root->next[i]=NULL;
        flag=0;
        for(int i=0;i<n;i++){
            scanf("%s",s);
            Trie(s);
        }
        if(flag) printf("NO\n");
        else     printf("YES\n");
        delTrie(root);
    }
    return 0;
}