POJ 1226 Substrings

Substrings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10262		Accepted: 3525

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：找出一组字符串中的最长公共子串或逆序子串的长度

#include <string>
#include <stdio.h>
#include <stdlib.h>
#include<iostream>  
#include<cstdio>  
#include<cstring>  
using namespace std;

char srcstr[110][110];//输入字符串
char str[110];//比较字符串
char strrev1[110];//翻转后的比较字符串
char strshort[110];//最短的字符串
int n;
int ncase;

void StrRev(char* pstr)
{
    int nLen = strlen(pstr);
    for (int i = nLen - 1; i >= 0; i--)
    {
        strrev1[nLen - i - 1] = pstr[i];
    }
    strrev1[nLen] = '\0';
}

int Find()
{
    int nLenshort = strlen(strshort);
    int nLen = nLenshort;
    while(nLen)
    {
        for (int j = 0; j  + nLen <= nLenshort; j++)
        {
            strncpy(str, strshort + j, nLen);
            str[nLen] = '\0';
            StrRev(str);
            for (int k = 0; k < n; k++)
            {
                if (!strstr(srcstr[k], str) && !strstr(srcstr[k], strrev1))
                {
                    break;
                }
                if (k == n - 1)
                {
                    return nLen;
                }
            }
        }
        nLen--;
    }
    return nLen;
}

int main()
{
    scanf("%d", &ncase);
    int MinLen = 1000;
    while(ncase--)
    {
        MinLen = 1000;
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
        {
            scanf("%s", srcstr[i]);
            if (MinLen > strlen(srcstr[i]))
            {
                MinLen = strlen(srcstr[i]);
                strcpy(strshort, srcstr[i]);
            }
        }
        printf("%d\n", Find());
    }
    return 0;
}