Substrings

Substrings

时间限制(普通/Java):3000MS/10000MS          运行内存限制:65536KByte
总提交:60            测试通过:32

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 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.

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

思路： 先找出最短的母串，即该符合要求的子串肯定在这个母串中，即在从长到短
从最短母串中取子串，在子串正反去查看是否符合要求。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    string s[200];
    int n,m,i,j,min1,minline,k;
    scanf("%d",&n);
    while(n--)
    {
        int max1=0;
        min1=10000;
        scanf("%d",&m);
        getchar();
        for(i=0; i<m; i++)
        {
            cin>>s[i];
            int len=s[i].size();
            if(len<min1)
            {
                min1=len;
                minline=i;
            }
        }
       // printf("%d\n",min1);
        for(i=s[minline].size(); i>=0; i--)
        {
            for(j=0; j<=i; j++)
            {
                string s1,s2;
                s1 = s[minline].substr(j, i);
                s2=s1;
                reverse(s2.begin(), s2.end());
                for(k=0; k<m; k++)
                {
                    if(s[k].find(s1)==-1&&s[k].find(s2)==-1)
                    {
                        break;
                    }
                }
                if(k==m&&max1<s1.size())
                {
                    max1=s1.size();
                }
            }
        }
        printf("%d\n",max1);
    }
    return 0;
}