poj 3630 Phone List trie树

Phone List

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:

• Emergency 911
• Alice 97 625 999
• 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

Nordic 2007

题意：如果有一个字符串是别的字符串的前缀就输出NO，否则YES；

思路：最基本的trie树应用；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e5+10,M=5e6+10,inf=1e9+10;
int a[N][27],sum[M],len;
void init()
{
    memset(a,0,sizeof(a));
    memset(sum,0,sizeof(sum));
    len=1;
}
int getnum(char a)
{
    return a-‘0‘;
}
void insertt(char *aa)
{
    int u=0,n=strlen(aa);
    for(int i=0; i<n; i++)
    {
        int num=getnum(aa[i]);
        if(!a[u][num])
        {
            a[u][num]=len++;
        }
        u=a[u][num];
        sum[u]++;
    }
}
int getans(char *aa)
{
    int u=0,x=strlen(aa);
    for(int i=0; i<x; i++)
    {
        int num=getnum(aa[i]);
        if(!a[u][num])
            return 0;
        u=a[u][num];
    }
    return sum[u];
}
char ch[N][15];
int main()
{
    int x,y,z,i,t;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&x);
        init();
        for(i=0; i<x; i++)
        {
            scanf("%s",ch[i]);
            insertt(ch[i]);
        }
        int flag=1;
        for(i=0;i<x;i++)
        {
            if(getans(ch[i])!=1)
            {
                flag=0;
                break;
            }
        }
        if(flag)
        printf("YES\n");
        else
        printf("NO\n");
    }
    return 0;
}