OpenJudge/Poj 1226 Substrings

1.链接地址：

http://bailian.openjudge.cn/practice/1226/

http://poj.org/problem?id=1226

2.题目：

总时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

The first line of the input 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.

输出

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

样例输入

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

样例输出

2
2 

来源

Tehran 2002 Preliminery

3.思路：

4.代码：

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string.h>
 using namespace std;
 const int NUM = ;
 char strs[NUM][NUM + ];
 int main()
 {
     //freopen("F:\\input.txt","r",stdin);
     int i,j,k;

     int t;
     cin>>t;

     int n,length;
     while(t--)
     {
         cin>>n;
         cin.get();

         for(i = ; i < n; i++)
         {
             scanf("%s",strs[i]);
         }

         for(i = ; i < n; i++)

         length = strlen(strs[]);
         char substr[NUM + ],substr2[NUM + ];
         int res = ;
         for(i = ; i <= length; i++)
         {
             for(j = ; (j+i-) < length; j++)
             {
                 strncpy(substr,&strs[][j],i);
                 substr[i] = '\0';
                 strcpy(substr2,substr);

                 for(k = ; k < (i+)/; k++)
                 {
                     char tmp = substr2[k];
                     substr2[k] = substr2[i--k];
                     substr2[i--k] = tmp;
                 }

                 //cout<<"substr="<<substr<<",substr2="<<substr2<<endl;
                 for(k = ; k < n; k++)
                 {
                     if(!strstr(strs[k],substr) && !strstr(strs[k],substr2)) break;
                 }
                 if(k >= n )
                 {
                     res = i;
                     break;
                 }
             }
         }

         cout<<res<<endl;
     }
     return ;
 }