codeforces Round #451 (Div. 2) Phone Numbers
2017-12-16 22:57
232 查看
思路:直接暴力就好了,写这道题主要还是熟悉一下对stl的用法。
每次插入map去重,然后每次双重循环判断是否是某个数的字串。
每次插入map去重,然后每次双重循环判断是否是某个数的字串。
#include <cstdio> #include <cstring> #include <iostream> #include <string> #include <set> #include <map> #include <vector> using namespace std; typedef long long LL; map<string, set<string> > mp; map<string, set<string> > ans; bool judge(string s1, string s2) { if(s1.size()>s2.size()) return false; for(int i=0;i<s1.size();++i) { if(s1[i]!=s2[s2.size()-s1.size()+i]) return false; } return true; } /* bool judge(string s1, string s2) { if(s1.size()>=s2.size()) return false; int x=s2.find(s1,s2.size()-s1.size()); //cout<<s1<<" "<<s2<<" "<<x<<endl; if(x==s2.size()-s1.size()) return true; else return false; } */ int main() { int n, num; string s, ss; ios::sync_with_stdio(false); cin>>n; for(int i=0;i<n;++i) { cin>>s>>num; for(int j=0;j<num;++j) { cin>>ss; mp[s].insert(ss); } } cout<<mp.size()<<endl; map<string, set<string> >::iterator it; set<string> ::iterator its, itss; for(it=mp.begin();it!=mp.end();++it) { for(its=(*it).second.begin();its!=(*it).second.end();++its) { string s1=*its; int flag=0; for(itss=(*it).second.begin();itss!=(*it).second.end();++itss) { string s2=*itss; if(itss==its) continue; if(judge(s1,s2)) { flag=1; break; //cout<<*its<<endl; } } if(!flag) { ans[it->first].insert(s1); } } } for(it=ans.begin();it!=ans.end();++it) { cout<<it->first<<" "<<it->second.size(); for(its=it->second.begin();its!=it->second.end();its++) { cout<<" "<<*its; } cout<<"\n"; } return 0; }
相关文章推荐
- Codeforces Round #160 (Div. 2)---A. Roma and Lucky Numbers
- Codeforces Round #FF (Div. 1) C. DZY Loves Fibonacci Numbers
- Valid Phone Numbers
- Codeforces Round #358 (Div. 2) A Alyona and Numbers(水题)
- Codeforces Round #361 (Div. 2) -- A. Mike and Cellphone (思路题目)
- Codeforces Round #452 (Div. 2) - C. Dividing the numbers (思路)
- Codeforces Round #155 (Div. 2) A. Cards with Numbers
- Codeforces Round #330 (Div. 2) B. Pasha and Phone 容斥定理
- Codeforces Round #232 (Div. 2) B - On Corruption and Numbers
- Codeforces Round #432 (Div. 2) D 850B Arpa and a list of numbers(gcd 枚举)
- Codeforces Round #136 (Div. 2) B. Little Elephant and Numbers
- Bestcoder #82 Div2 ztr loves lucky numbers(next_permutation)
- 193 Valid Phone Numbers
- Codeforces Round #358 (Div. 2) A. Alyona and Numbers
- timus 1002. Phone Numbers(KMP&动态规划)
- Codeforces Round #361 (Div. 2)A. Mike and Cellphone
- Codeforces Round #358 (Div. 2)(A) Alyona and Numbers
- CodeForces Round #112 Div2 165 E. Compatible Numbers
- Codeforces Round #452 (Div. 2) C - Dividing the numbers 推规律
- 193. Valid Phone Numbers