学步园

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

Author
Asia 2002, Tehran (Iran), Preliminary

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先二分长度,然后枚举这样长度的子串,依次去匹配

/*************************************************************************
    > File Name: hdu1238.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月17日 星期二 15时07分41秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 110;
int next[N];
char str[N];
char S[N][N];
char T[N];

void get_next ()
{
    next[0] = -1;
    int j = 0;
    int k = -1;
    int len = strlen (str);
    while (j < len)
    {
        if (k == -1 || str[j] == str[k])
        {
            next[++j] = ++k;
        }
        else
        {
            k = next[k];
        }
    }
}

bool MATCH ()
{
    int len1 = strlen (T);
    int len2 = strlen (str);
    int i = 0;
    int j = 0;
    while (i < len1 && j < len2)
    {
        if (j == -1 || T[i] == str[j])
        {
            ++i;
            ++j;
        }
        else
        {
            j = next[j];
        }
    }
    return j == len2;
}

void BinSearch (int n)
{
    int l = 1;
    int r = strlen (S[1]);
    int len = r;
    int mid;
    int ans = 0;
    while (l <= r)
    {
        mid = (l + r) >> 1;
        bool G = false;
        for (int i = 0; i <= len - mid; ++i)
        {
            for (int j = i; j < i + mid; ++j)
            {
                str[j - i] = S[1][j];
            }
            str[mid] = '\0';
            get_next ();
            bool flag = true;
            for (int j = 2; j <= n; ++j)
            {
                strcpy (T, S[j]);
                if (MATCH())
                {
                    continue;
                }
                int x = strlen (T);
                for (int k = 0; k < x / 2; ++k)
                {
                    swap (T[k], T[x - k - 1]);
                }
                if (MATCH())
                {
                    continue;
                }
                else
                {
                    flag = false;
                    break;
                }
            }
            if (flag)
            {
                G = true;
                break;
            }
        }
        if (G)
        {
            l = mid + 1;
            ans = mid;
        }
        else
        {
            r = mid - 1;
        }
    }
    printf("%d\n", ans);
}

int main ()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int n;
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i)
        {
            scanf("%s", S[i]);
        }
        BinSearch(n);
    }
    return 0;
}