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 = 100;
char strs[NUM][NUM + 1];
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 = 0; i < n; i++)
        {
            scanf("%s",strs[i]);
        }
        
        for(i = 0; i < n; i++)

        length = strlen(strs[0]);
        char substr[NUM + 1],substr2[NUM + 1];
        int res = 0;
        for(i = 1; i <= length; i++)
        {
            for(j = 0; (j+i-1) < length; j++)
            {
                strncpy(substr,&strs[0][j],i);
                substr[i] = '';
                strcpy(substr2,substr);
                
                for(k = 0; k < (i+1)/2; k++)
                {
                    char tmp = substr2[k];
                    substr2[k] = substr2[i-1-k];
                    substr2[i-1-k] = tmp;
                }
                
                
                //cout<<"substr="<<substr<<",substr2="<<substr2<<endl;
                for(k = 1; k < n; k++)
                {
                    if(!strstr(strs[k],substr) && !strstr(strs[k],substr2)) break;
                }
                if(k >= n ) 
                {
                    res = i;
                    break;
                }
            }
        }
        
        cout<<res<<endl;
    }
    return 0;
}