HDU1671-Phone List

Phone List

                                                                             Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                      Total Submission(s): 18182    Accepted Submission(s): 6107

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）

 

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字典树：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>using namespace std;int cmp(string a,string b){    return a>b;}int f[100009][15];int main(){    int t,n;    string s[10009];    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)            cin>>s[i];        sort(s,s+n,cmp);        memset(f[0],0,sizeof f[0]);        int sum=0,flag1=1;        for(int i=0;i<n;i++)        {            int flag2=1,Size=s[i].size();            for(int j=0,k=0;j<Size;j++)            {                int x=s[i][j]-'0'+1;                if(!f[k][x])                {                    f[k][x]=++sum;                    flag2=0;                    memset(f[sum],0,sizeof f[sum]);                }                k=f[k][x];            }            if(flag2) flag1=0;        }        if(!flag1) printf("NO\n");        else printf("YES\n");    }    return 0;}

暴力：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>using namespace std;struct node{    char s[20];    friend bool operator<(node x,node y)    {        return strcmp(x.s,y.s)>0;    }}a[10009];int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%s",a[i].s);        sort(a,a+n);        int flag=1;        for(int i=0;i<n-1;i++)        {            if(strstr(a[i].s,a[i+1].s))            {                flag=0;break;            }        }        if(!flag) printf("NO\n");        else printf("YES\n");    }    return 0;}