Substrings--poj1226(字符串)

Posted 2020-06-09

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 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：
给你n个字符串  要你找这n个字符串中最大的相同字串（字符串可以逆置）

分析：
先找到最小的母串  然后这个母串的所有字串从大到小 都和n个串循环一遍  只要找到符合的答案就跳出

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#define N 220
#define INF 0xffffffff
using namespace std;

int main()
{
    int T,n;
    char str[N][N],s1[N],s2[N];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int Min=INF,b;
        for(int i=0; i<n; i++)
        {
            scanf("%s",str[i]);
            if(strlen(str[i])<Min)
            {
                Min=strlen(str[i]);
                b=i;
            }
        }
        int len=Min,flag;
        while(len)
        {
            for(int i=0; i<=Min-len; i++)
            {
                flag=0;
                strncpy(s1,str[b]+i,len);
                int k=0;
                for(int i=len-1; i>=0; i--)
                {
                    s2[k++]=s1[i];
                }
                s1[len]=s2[len]=‘\0‘;
                for(int i=0; i<n; i++)
                {
                    if(strstr(str[i],s1)==NULL && strstr(str[i],s2)==NULL)
                    {
                        flag=1;
                        break;
                    }
                }
                if(flag==0)
                {
                    break;
                }
            }
            if(flag==0)
                break;
            len--;
        }
        printf("%d\n",len);
    }
    return 0;
}