Substrings

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

 

题目大意：求最长公共子串，正反都可以，可以直接暴力来求；

需要用到一些字符串函数，在代码中给出解释

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N=105;

int len[N];
char s[N][N];
char s1[N],s2[N];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int j,i,n,mini=inf,u;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%s",s[i]);
            len[i]=strlen(s[i]);
            if(len[i]<mini)//记录最短的主串，
            {              //从最短的主串中的每个字串找
                mini=len[i];
                u=i;
            }
        }
        int l,maxx=0,k,g;
        for(i=0;i<mini;i++)
        {
            for(j=1;j<=mini-i;j++)
            {
                strncpy(s1,s[u]+i,j);//strncpy(s1,s[u]+i,j)，作用是
                s1[j]='\0';          //复制s[u]从i开始以后的j个字符到s1中
                l=strlen(s1);
                for(k=l-1,g=0;k>=0;k--)
                    s2[g++]=s1[k];
                s2[l]='\0';
                int flag=1;
                for(k=0;k<n;k++)
                {
                    if(!strstr(s[k],s1)&&!strstr(s[k],s2))
                    {//strstr(s,s1);作用是判断s中是否出现s1，出现返回true，没有返回false；
                        flag=0;
                        break;
                    }
                }
                if(flag==1)
                    maxx=max(maxx,l);
            }
        }
        printf("%d\n",maxx);
    }
    return 0;
}