HDU1671：Phone List（字典树）

来源：互联网发布：php短链接生成器源码编辑：程序博客网时间：2024/04/30 04:21

Phone List

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19457 Accepted Submission(s): 6568

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）

题意：给出N个电话号码，若果其中一个号码为另一个号码的前缀，就输出NO，否则输出YES。

思路：如果a是b的前缀，当a比b先进入字典树时，处理b容易发现a已经存在。当b比a先进入字典树时，处理a时加个mark变量记录a有无产生新节点，无就证明以a为前缀的号码已经存在。另一种方法是用string存号码，按长度小到大排序，就能肯定短的先进入字典树了，不过这样做效率较低。

# include <iostream># include <cstdio># include <algorithm># include <cstring>using namespace std;int cnt, flag;char s[10005][13];struct node{    int next[10], tag;}tri[100050];void add(char *s){    int pos = 0, mark = 0, len = strlen(s);    for(int i=0; i<len; ++i)    {        int t = s[i]-'0';        if(!tri[pos].next[t])        {            mark = 1;            tri[pos].next[t] = ++cnt;        }        pos = tri[pos].next[t];        if(tri[pos].tag)        {            flag = 1;            break;        }    }    tri[pos].tag = 1;    if(!mark)        flag = 1;}int main(){    int t, n;    scanf("%d",&t);    while(t--)    {        flag = cnt = 0;        scanf("%d",&n);        memset(tri, 0, sizeof(tri));        for(int i=0; i<n; ++i)            scanf("%s",s[i]);        for(int i=0; i<n; ++i)            if(!flag)                add(s[i]);        if(flag)            puts("NO");        else            puts("YES");    }}

0 0