HDU1238 Substrings

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6185    Accepted Submission(s): 2747

Problem Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

 

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

 

Output

There should be one line per test case containing the length of the largest string found.

 

Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

 

Sample Output

2
2

 

Author

Asia 2002, Tehran (Iran), Preliminary

 

Recommend

Eddy

代码如下：

#include<iostream>
#include<string>
using namespace std;
int n,ans;
string str[110];
void dfs(string temp,string retemp,int cur){
    if(temp.size()<=ans)  return;
    if(cur==n&&temp.size()>ans)
    {
        ans=temp.size(); return;
    }
    if(str[cur].size()<temp.size()) return ;
    string now_str="";
    for(int i=0;i<=(str[cur].size()-temp.size());i++){
        now_str=str[cur].substr(i,temp.size());
        if(now_str==temp||now_str==retemp){

            dfs(temp,retemp,cur+1);
            break;
        }
    }
    return ;

}
int main(){
    int t,i,j,k;
    string temp="",retemp="";
    cin>>t;
    while(t--){
        cin>>n;
        for(i=0;i<n;i++)
        cin>>str[i];
        ans=0;

        for(i=0;i<str[0].size();i++){
            temp=""; retemp="";
            for(j=i;j<str[0].size();j++)
            {
                temp+=str[0][j]; retemp="";
                for(k=temp.size()-1;k>=0;k--)
                   retemp+=temp[k];
           //    cout<<temp<<" "<<retemp<<endl;
                dfs(temp,retemp,1);
            }

        }
        cout<<ans<<endl;


    }

}