Phone List

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

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20211    Accepted Submission(s): 6836

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

 方法一：字典树（注意字符串最后的判断）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
struct trie
{
    bool isend;
    struct trie *next[10];
};
trie *root;
bool flag;
void buildtrie(trie *root,char *s)
{
    trie *p=root;
    trie *tmp;
    int len=strlen(s);
    for(int i=0;i<len;i++)
    {
        if(p->next[(int)(s[i]-'0')]==NULL){
            tmp=new trie();
            for(int j=0;j<10;j++)
                tmp->next[j]=NULL;
            tmp->isend=false;
            p->next[(int)(s[i]-'0')]=tmp;
        }
        else
        {
            if(p->next[s[i]-'0']->isend==true)
            {
                flag=false;
                return;
            }
            else if(i==len-1 && p->next[s[i]-'0']!=NULL)
            {
                flag=false;
                return ;
                /*
                测试数据
                1
                3
                97625999
                91125426
                911
                 */
            }
        }
        p=p->next[s[i]-'0'];
    }
    p->isend=true;
}
void del(trie *root)
{
    if(root==NULL) return;
    for(int i=0;i<10;i++)
    {
        if(root->next[i]!=NULL)
            del(root->next[i]);
    }
    free(root);
}
int main()
{
    int t,n;
    char a[15];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int x=0;
        root=new trie();
        root->isend=false;
        for(int i=0;i<10;i++)
            root->next[i]=NULL;
        flag=true;
        for(int i=0;i<n;i++)
        {
            scanf("%s",a);
            if(flag==true)buildtrie(root,a);
        }
        if(flag==false) printf("NO
");
        else printf("YES
");
        del(root);
    }
    return 0;
}

方法二：放入容器内排序查找

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
int main()
{
    vector<string>v;
    char a[10];
    string s;
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        cin>>n;
        v.clear();
        for(int i=0;i<n;i++)
        {
            scanf("%s",a);
            s=a;
            v.push_back(s);
        }
        sort(v.begin(),v.end());
        for(int i=0;i<n-1;i++)
        {
            if(v[i+1].find(v[i])==0)
            {
                printf("NO
");
                goto eg;
            }
        }
        printf("YES
");
        eg:;
    }
    return 0;
}