Hdu 1671 Phone List 字典树+变形

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15867    Accepted Submission(s): 5336

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

2391197625999911254265113123401234401234598346

 

Sample Output

NOYES

 

Source

2008 “Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（3）

 

  思路：简单题目，字典树应用+变形，主要是要注意一点，合法的号码是没有前缀也不是其他号码的前缀，也就是既要判断是不是他人的前缀，又要判断他人是不是自己的前缀，加两个标志就行了。

  AC代码如下：

#include <iostream>#include <cstring>#include <cstdio>using namespace std;struct Node{    Node *next[15];    bool fsign;    bool nsign;    Node(){        memset(next,NULL,sizeof(next));        fsign=nsign=false;    }}*root;Node * getNode(){    Node * tmp=new Node();    return tmp;}void Insert(char *p){    Node * tmp=root;    while(*p){        int id=*p-'0';        if(tmp->next[id]==NULL){            tmp->next[id]=getNode();            tmp->fsign=true;        }        tmp=tmp->next[id];        p++;    }    tmp->nsign=true;}bool Search(char *p){    Node *tmp=root;    while(*p){        int id=*p-'0';        if(tmp->nsign) return true;        if(tmp->next[id]!=NULL){            tmp=tmp->next[id];            p++;        }        else return false;    }    if(tmp->fsign) return true;    else return false;}void del(Node * root){    for(int i=0;i<15;i++){        if(root->next[i])            del(root->next[i]);    }    delete(root);}int main(){    int t,n;    char tmp[15];    bool ok;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        ok=false;        root=getNode();        while(n--){            scanf("%s",tmp);            if(Search(tmp)){                ok=true;            }            if(ok) continue;            Insert(tmp);        }        if(ok) printf("NO\n");        else printf("YES\n");        del(root);    }    return 0;}