《ACM程序设计》书中题目X

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原创最新推荐文章于 2026-06-12 10:09:26 发布 448 阅读 AI 写同款· GEO 优化 ›

ACM程序设计学期总结心路历程大一的时候就跟着费老师学了近一个学期的ACM程序设计课程，但是由于种种出自于自己的原因，当时并没有完全塌下心去学，只是学到了些基础的东西，做的题太少，当然也并没有深入理解。大二又有了这个机会，自然不能放过，又认认真真重新学了一遍。虽然这学期这门课一共只学了四个专题，但是对我来说，收获非常大，正像费老师说的：acm是思维的体操。它不停留在基本的语言层面，而是专注于解决各种实际问题的有阅读详情

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
X不是Z的前缀，并且Z的字典顺序比X大，那么只存在两种情况:
1)n>=t，这种情况下，必定存在一个i(0<=i<=t)使得X[i]

#include<bits/stdc++.h>
using namespace std;
char phone[10000][11];
int cmp(const void* a,const void* b)
{
    return strcmp((char*)a,(char*)b);
}

int main(void)
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int i,n;
        bool flag=false;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%s",phone[i]);
        }
        qsort(phone,n,sizeof(phone[0]),cmp);
        for(i=0;i<n-1;i++)
        {
            if(strncmp(phone[i],phone[i+1],strlen(phone[i]))==0)
            {
                flag=true;
                break;
            }
        }
        if(flag==false)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}