The first ACM College student Program Design competition in Shandong Province Phone number Dictionary tree

Phone Number Time limit:1000ms Memory limit:65536k have questions. Dot here ^_^ Topic DescriptionWe know that if a phone number A is another phone number B's prefix, B is isn't able to be called. For a example, A are 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's A is B ' s prefix.

inputThe input consists of several test cases.
The "a" 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 are followed by a line containing one zero. OutputFor each test case, if there exits a phone number that is cannot be called, print "NO", otherwise print "YES" instead. Sample Input

2
012
012345
2
012345
0

Sample Output

NO
YES

Tips SourceThe first ACM College student Program Design competition in Shandong Province in 2010

Water Problem dictionary tree to determine if there is a prefix number appears

Accode:

#include <map> #include <queue> #include <cmath> #include <cstdio> #include <cstring> # Include <stdlib.h> #include <iostream> #include <algorithm> #define MAX #define MAXN 1005 using names
Pace 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;
 }/* 2 012 012345 2 12 012345 0 * *

Violent practices

#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include < cmath> #include <cstring> #include <string> #include <vector> #define MAXN 1005 #define INF 0X7FFFFFF
FFF #define LL-long long using namespace std;
int n,m,loop;
    struct N {char S[MAXN];
int Len;
}MY[MAXN];
BOOL CMP (N A,n b) {return a.len<b.len;} int LEN[MAXN];
            int main () {while (scanf ("%d", &n) &&n) {for (int i=1;i<=n;++i) {scanf ("%s", MY[I].S);
        My[i].len=strlen (MY[I].S);
        Sort (my+1,my+1+n,cmp);
        BOOL Flag=true;
                for (int i=1;flag&&i<=n;++i) for (int j=i+1;flag&&j<=n;++j) {int ok=0;
                for (int z=0;z<my[i].len;++z) if (my[i].s[z]==my[j].s[z]) ok++;
            if (Ok==my[i].len) Flag=false;
        if (flag) puts ("YES");
    Else puts ("NO");
return 0; }