HDU6208 The Dominator of Strings 字符串匹配 2017 ACM/ICPC Asia Regional Qingdao Online
2017-09-18 20:10
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题目链接:HDU6208
题目思路:题意比较好懂,找到一个长度最长的串,然后判断其他所有串是否能否被这个串包含,也就是是否能够匹配成功,如果成功输出该串,否则输出No;
当时队友用字典树,AC自动机不是MLE 就是TLE 然后刚刚队友突然说,暴力也可以做……我就写了一发暴力……结果还真能过……(以后遇到不会的题目,先暴力再说)
AC
4000
code:
题目思路:题意比较好懂,找到一个长度最长的串,然后判断其他所有串是否能否被这个串包含,也就是是否能够匹配成功,如果成功输出该串,否则输出No;
当时队友用字典树,AC自动机不是MLE 就是TLE 然后刚刚队友突然说,暴力也可以做……我就写了一发暴力……结果还真能过……(以后遇到不会的题目,先暴力再说)
AC
4000
code:
/* 2017年9月18日20:08:05 HDU6208 AC */ #include <iostream> #include <map> #include <set> #include <string> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> #include <queue> #include <vector> using namespace std; const int maxn=1e5+10; string str[maxn]; int main(){ int t; ios::sync_with_stdio(0); //scanf("%d",&t); cin>>t; while(t--){ int n,mx=0; string p; cin>>n; for(int i=1;i<=n;i++){ cin>>str[i]; if(str[i].length()>mx) { mx=str[i].length(); p=str[i]; } } bool f=true; for(int i=1;i<=n;i++){ if(p.find(str[i])==-1){ f=false; break; } } if(!f) cout<<"No"<<endl; else cout<<p<<endl; } return 0; }
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