x-24

2019-11-06 06:30:06

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來源：轉載

供稿：網友

題目描述：DescriptionGiven 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 26In 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.InputThe 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.OutputFor each test case, output "YES" if the list is consistent, or "NO" otherwise.Sample Input2391197625999911254265113123401234401234598346Sample OutputNOYES簡單概述：即給你幾個字符串，讓你尋找是否有包含的關系，即某一字符串是不是為另一字符串的子串。解題思路：先將給出的字符串進行排序，然后尋找是否有包含的，注意運用標志flag和采用find函數。細節處理：由于習慣性問題，將尋找過程中的j寫成了i,導致不能AC,并且出現segmentation fault的錯誤。

代碼:

#include<bits/stdc++.h>using namespace std;int main(){	vector<string>v;	string s;	int m,n,i,j,k;	cin>>n;	for(i=0;i<n;i++)	{		int flag=0;		v.clear();		cin>>m;		for(j=0;j<m;j++)		{			cin>>s;			v.push_back(s);//壓入向量中		}		sort(v.begin(),v.end());		for(j=0;j<m-1;j++)		{			if(v[j+1].find(v[j])==0)//尋找是否包含子串			{				flag++;//建立的標志				}			}			 if(flag==0)			 cout<<"YES"<<endl;			 else       cout<<"NO"<<endl;		}			return 0;	}心得體會：
做題應先想清楚在敲代碼，注意一定要仔細！避免低級性錯誤！

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