http://acm.hdu.edu.cn/showproblem.php?pid=2328

Corporate Identity

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 698    Accepted Submission(s): 281

Problem Description

Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a help with their new identity. IBM do not want to change their existing logos and trademarks completely, because their customers are used to the old ones. Therefore, ACM will only change existing trademarks instead of creating new ones.

After several other proposals, it was decided to take all existing trademarks and find the longest common sequence of letters that is contained in all of them. This sequence will be graphically emphasized to form a new logo. Then, the old trademarks may still be used while showing the new identity.

Your task is to find such a sequence.

Input

The input contains several tasks. Each task begins with a line containing a positive integer N, the number of trademarks (2 ≤ N ≤ 4000). The number is followed by N lines, each containing one trademark. Trademarks will be composed only from lowercase letters, the length of each trademark will be at least 1 and at most 200 characters.

After the last trademark, the next task begins. The last task is followed by a line containing zero.

Output

For each task, output a single line containing the longest string contained as a substring in all trademarks. If there are several strings of the same length, print the one that is lexicographically smallest. If there is no such non-empty string, output the words “IDENTITY LOST” instead.

Sample Input

3

aabbaabb

abbababb

bbbbbabb

2

xyz

abc

0

Sample Output

abb
IDENTITY LOST

其实就是一个暴力解决的问题，以我现在的程度只能暴力，可是我也想写个更好的方法呀，下次一定要学

可坑了，我没有给我的存子串的串赋初值，导致我在用ｓｔｒｃｍｐ的时候老是出错，弄的我稀里糊涂的写了个Ｊｕｄｇｅ函数用来比较串，可道理是一样的呀，于是又ｗａ了，还好在队友的帮助下找到了，初值这个问题不知道害了我多少次，可我还往里面跳，以后要更加注意了，有的错找不出来了，可以往初值上想想，另外以后要注意赋初值

代码：

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

#define M 40005
#define N 210

char s[M][N];
int Next[N];

void FindNext(char b[])
{
    int i=0, j=-1, blen=strlen(b);
    Next[0] = -1;

    while(i<blen)
    {
        if(j==-1 || b[i]==b[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}

int KMP(char a[], char b[])
{
    int i=0, j=0;
    int alen=strlen(a), blen=strlen(b);

    FindNext(b);

    while(i<alen)
    {
        while(j==-1 || (a[i]==b[j] && i<alen && j<blen))
            i++, j++;
        if(j==blen)
            return 1;
        j = Next[j];
    }
    return 0;
}

int main()
{
    int n;
    while(scanf("%d", &n), n)
    {
        int i, j, k, MinLen=1000, len;
        char ss[N];

        memset(s, 0, sizeof(s));

        for(i=0; i<n; i++)
        {
            scanf("%s", s[i]);
            len = strlen(s[i]);

            if(len<MinLen)
            {
                MinLen = len;
                memset(ss, 0, sizeof(ss));
                strcpy(ss, s[i]);
            }
        }

        char b[N]="{";
        int index=0;

        for(i=MinLen; i>0; i--)
        {
            for(j=0; j<=MinLen-i; j++)
            {
                char a[N];

                memset(a, 0, sizeof(a));

                strncpy(a, ss+j, i);

                for(k=0; k<n; k++)
                {
                    if(KMP(s[k], a)==0)
                        break;
                }

                if(k==n && strcmp(a, b)<0)
                {
                    index = i;
                    strcpy(b, a);
                }

                if(index && j==MinLen-i)
                    i=-1, j=1000;
            }
        }

        if(index)
            printf("%s\n", b);
        else
            printf("IDENTITY LOST\n");

    }
    return 0;
}