HDU1238：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：找出所有串的最长的公共连续子串

思路：直接从最小的那串，枚举所有子串去寻找，反正最多100串，最长100字符，大胆的枚举吧！骚年！

 

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

int main()
{
    int t,n,i,j,k,MIN,f,len,MAX;
    char str[105][105],s1[105],s2[105];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        MIN = 1000;
        for(i = 0; i<n; i++)
        {
            scanf("%s",str[i]);
            len = strlen(str[i]);
            if(MIN>len)//找到最小串
            {
                MIN = len;
                f = i;
            }
        }
        len = strlen(str[f]);
        int flag = 1;
        MAX = 0;
        for(i = 0;i<len;i++)//作为标本串子串的头
        {
            for(j = i;j<len;j++)//子串的尾
            {
                for(k = i;k<=j;k++)//复制为两个串，顺序串s1，逆序串s2
                {
                    s1[k-i] = str[f][k];
                    s2[j-k] = str[f][k];
                }
                s1[j-i+1] = s2[j-i+1] = '\0';
                int l = strlen(s1);
                for(k = 0;k<n;k++)//枚举所有串
                {
                    if(!strstr(str[k],s1) && !strstr(str[k],s2))
                    {
                        flag = 0;
                        break;
                    }
                }
                if(l>MAX && flag)
                MAX = l;
                flag = 1;
            }
        }
        printf("%d\n",MAX);
    }

    return 0;
}