山东省第一届ACM大学生程序设计竞赛 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 integerN (0<N<1001), represent the number of phone numbers.
 The next line containsN 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

20120123452120123450

示例输出

NOYES

提示

 

来源

 2010年山东省第一届ACM大学生程序设计竞赛

水题字典树，判断是否有前缀号码出现

ACcode：

#include <map>#include <queue>#include <cmath>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define MAX 10#define maxn 1005using namespace std;struct Trie{    Trie* next[MAX];    int v;};struct STR{    char str[maxn];    int len;    bool operator<(const STR &a)const{        return (this->len<a.len);    }}my[maxn];Trie *root;int n,ans,num;void createTire(char *str){    int len=strlen(str);    Trie *p=root,*q;    for(int i=0;i<len;++i){        int id=str[i]-'0';        if(p->next[id]==NULL){            q=(Trie *)malloc(sizeof(Trie));            q->v=1;            for(int j=0;j<MAX;++j)                q->next[j]=NULL;            p->next[id]=q;            p=p->next[id];        }        else{            p->next[id]->v++;            p=p->next[id];        }    }    p->v=-1;}int findTrie(char *str){    int len=strlen(str);    Trie *p=root;    for(int i=0;i<len;++i){        int id=str[i]-'0';        p=p->next[id];        if(p==NULL)            return 0;        if(p->v==-1)            return 0;    }    return 2;}void init(){    root=(Trie *)malloc(sizeof(Trie));    for(int i=0;i<MAX;++i)        root->next[i]=NULL;    ans=num=0;    memset(my,0,sizeof(my));}int main(){    while(~scanf("%d",&n)&&n){        init();        for(int i=0;i<n;++i){            scanf("%s",my[i].str);            my[i].len=strlen(my[i].str);        }        sort(my,my+n);        for(int i=0;i<n;++i)createTire(my[i].str);     //   for(int i=0;i<n;i++)cout<<my[i].str<<'\12';        for(int i=0;i<n;++i)            ans+=findTrie(my[i].str);            //cout<<"ans: "<<ans<<'\12';        for(int i=0;i<n&&!ans;++i)            for(int j=i+1;j<n;++j)                if(!strcmp(my[i].str,my[j].str))                    ans++;        printf(ans>0?"NO\n":"YES\n");    }    return 0;}/*20120123452120123450*/