Substrings


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9246    Accepted Submission(s): 4388

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


题解：找最长的公共子串，反串的子串也算。注意：子串是连续的，子序列是不用连续的。


AC代码：DFS


#include <iostream>
#include<stdio.h>
#include <algorithm>
using namespace std;
string str[110]; 
int t;
int cmp(const string& a, const string& b){
  return a.length()<b.length();
}
//string substr(int pos = 0,int n = npos) const;  返回从pos开始的n个字符组成的字符串

int dfs(string now,string reve,int len)
{
  for(int i=1;i<t;i++)
  {
    int sign=0;
    for(int j=0;j+len<=str[i].length();j++)
    {
      string sub=str[i].substr(j,len);  //返回从j开始 ,长度为len的子串
      
      if(sub==now||sub==reve) 
      {
        sign=1;break; //找到就继续找
      }
    }
    if(!sign)
      return 0; //否则直接退出
  }
  return len;
}
int main(){
  int n;
  cin>>n;
  while(n--){
    cin>>t;
    for(int i=0;i<t;i++)
      cin>>str[i];
    sort(str,str+t,cmp);
    int len=str[0].length();
    int max=0;
    for(int i=len;i>=1;i--)
    {
      for(int j=0;j+i<=len;j++)
      {
        string now=str[0].substr(j,i); //回从 j开始的 i个字符组成的字符串
        string reve=now;
        reverse(reve.begin(),reve.end());
        //手写的翻转字符串,与上一行效果相同 
        //for(int k=now.length()-1;k>=0;k--) 
        //  rev+=now[k];
        //cout<<rev<<endl;
        int maxn=dfs(now,reve,i);
        if(maxn>max)
          max=maxn;
      }
    }
    cout<<max<<endl;
  }
  return 0;
}


暴力：


#include<iostream>
#include<stdio.h >
#include<string>
#include<algorithm>
using namespace std;
string a[110];
int main()
{
  string s1,s2;
  int t,n,i,j,mini,min,k,max;
  cin>>t;
  while(t--)
  {
    max = 0; min = 200;
    cin>>n;
    for(i=0; i<n; i++)
    {
      cin>>a[i];
      if(a[i].size() < min)
      {
        min = a[i].size();
        mini = i;
      }
    }
    for(i=0; i<a[mini].size(); i++)
    {
      for(j=1; j<=a[mini].size() - i; j++)
      {
        s1 = a[mini].substr(i,j); //返回从i开始的 j 个字符组成的字符串
        s2 = s1;
        reverse(s2.begin(),s2.end());
        for(k=0; k<n; k++)
        {
          //从0开始查找字符串s1，s2 在当前串中的位置，查找成功时返回所在位置，失败返回string::npos的值
          if(a[k].find(s1,0) == string::npos && a[k].find(s2,0) == string::npos)  
            break;
        }
        if(k == n && j > max)
          max = j;
      }
    }
    cout<<max<<endl;
  }
  return 0;
}