POJ 1238 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

题意：n个字符串，找出这n个字符串中连续相同的最大子序列，子串的逆序也可以。

#include<stdio.h>
#include<string.h>

const int N=110;

char str[N][N], resouce[N];
int n;

void Init() ///在输入数据时找到最短的那条字符串，只有这样才能保证找到的子串有可能是公共子串
{
    int i;

    scanf("%d", &n);
    scanf("%s", str[0]);

    strcpy(resouce, str[0]);

    for (i = 1; i < n; i++)
    {
        scanf("%s", str[i]);
        if (strlen(str[i]) < strlen(resouce))
            strcpy(resouce, str[i]);
    }
}

int Judge(char sub[], char resub[]) ///判断找到的子串和其逆序是否是其它字符串的子串
{
    int i;

    for (i = 0; i < n; i++)
    {
        if (strstr(str[i], sub) == NULL && strstr(str[i], resub) == NULL)
            return 0; ///只要在一个字符串中没有找到该子串和其逆序，则该子串不是公共子序列
    }

    return 1;
}

int Sub()
{
    int i, j, len1 = strlen(resouce), k, t;
    char sub[N], resub[N];

    for (i = len1; i >= 1; i--)
    {
        for (j = 0; i+j <= len1; j++) ///从最长的子串查找，只要找到了属于所有字符串的子串，最长公共子序列就是该子串，便可停止查找
        {
            memset(sub, 0, sizeof(sub)); ///最短字符串的子串
            memset(resub, 0, sizeof(resub)); ///子串的逆序
            k = 0;

            strncpy(sub, resouce+j, i);
            for (t = strlen(sub)-1; t >= 0; t--)
                resub[k++] = sub[t];
            resub[k] = ‘\0‘;

            if (Judge(sub, resub))
                return strlen(sub);
        }
    }

    return 0;
}

int main ()
{
    int T, len;

    scanf("%d", &T);

    while(T--)
    {
        memset(resouce, 0, sizeof(resouce));
        Init();

        len = Sub();

        printf("%d\n", len);
    }

    return 0;
}