（KMP 字符串处理）Substrings -- hdu -- 1238

http://acm.hdu.edu.cn/showproblem.php?pid=1238

 

Substrings

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1238

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

 

之前写个是完全的暴力写的，因为数据很小，暴力也没关系，这次用了KMP，有点小题大做的感觉，不过刚刚学，就当练习了。

 

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

#define MOD 10000
#define N 210

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

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

        scanf("%d", &n);

        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 a[N], b[N];
        int index=0;

        for(i=MinLen; i>0; i--)
        for(j=0; j<=MinLen-i; j++)
        {
            memset(a, 0, sizeof(a));
            memset(b, 0, sizeof(b));
            strncpy(a, ss+j, i);
            strcpy(b, a);
            strrev(b);

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

            if(k==n)
            {
                index = i;
                i=-1, j=1000;
            }
        }

        if(index)
            printf("%d\n", index);
        else
            printf("0\n");

    }
    return 0;
}