HDOJ1238(Substrings) 暴力过

Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1258    Accepted Submission(s): 572


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
//1245883 2009-04-08 08:52:09 Accepted 1238 15MS 248K 1234 B C++ Xredman 
#include <iostream>
#include 
<cstring>
using namespace std;

const int N = 102;

char str[N][N], shortStr[N];
int len[N], shortLen;
int t, n, cc;

bool found(char *pp)
{
    
for(int i = 0; i < n; i++)
        
if(i != cc && !strstr(str[i], pp) )
            
return false;
    
return true;
}


int main()

    
char ss[N], rr[N];
    
int tt, i, j, k;
    
bool flag;
    
while(cin>>t)
        
while(t-- && cin>>n)
        
{
            cc 
= shortLen = -1;
            
for(i = 0; i < n; i++)
            
{
                cin
>>str[i];
                len[i] 
= strlen(str[i]);
                
if(shortLen == -1 || len[i] < shortLen)
                
{
                    cc 
= i;
                    shortLen 
= len[i];
                }

            }

            strcpy(shortStr, str[cc]);
            tt 
= shortLen;flag = false;

            
while(tt > 0)
            
{
                
for(i = 0; i + tt <= shortLen; i++)
                
{
                    
for(j = i, k = 0; k < tt; k++, j++)
                        ss[k] 
= shortStr[j];
                    ss[k] 
= '\0';
                    
//cout<<"SS "<<ss<<endl;
                    if(found(ss))
                    
{
                        flag 
= true;
                        
break;
                    }

                    
for(j = shortLen - i - 1, k = 0; k < tt; k++, j--)
                        rr[k] 
= shortStr[j];
                    rr[k] 
= '\0';
                    
//cout<<"RR "<<rr<<endl;
                    if(found(rr))
                    
{
                        flag 
= true;
                        
break;
                    }

                }

                
if(flag)
                    
break;
                tt
--;
            }

            cout
<<tt<<endl;
        }

    
return 0;
}

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