HDU-1238 Substrings

                               Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6238    Accepted Submission(s): 2767

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

 

Author

Asia 2002, Tehran (Iran), Preliminary

 

Recommend

Eddy

//寻找最长字串并输出其长度
//  思路：
//  1、先寻找其中最短字符串str，求子字符串就要先找最短的字符串
//  2、根据str搜索满足条件的子字符串
// 3、对str的各子字符串从长到短一次判断是否满足条件，直到找到一个符合条件的子字符串为止
//strlen：计算字符串长度
//strstr：在字符串中寻找子字符串,它的用法用的少，但是碰到了，就要熟练。
//strncpy：复制字符串中的字串，它的用法要掌握。
//strcpy：复制字符串
//怎样找最短字符串，也是值得学习的。。
//搜索满足条件的最长子串.
#include<stdio.h>
#include<string.h>
char a[101][101];
int n;
void cmp(char *str)//字符串反序
{
    int l,i,g=0;
    char b[101];
    l=strlen(str);
    for(i=0;i<l;i++)
        b[g++]=str[l-i-1];
    b[g]='\0';
    strcpy(str,b);
}
int search(char *str)//搜索满足条件的最长子串并返回其长度
{
    int  l1,l2,i,j,f;
    char s[101],r[101];
    l1=strlen(str);l2=strlen(str);
      while(l1>0)//搜索不同长度的子串，从最长的子串开始搜索
      {
          for(i=0;i<=l2-l1;i++)
          {
              strncpy(s,str+i,l1);//strncpy的函数用法必须是str+i不能写成str【i】；
              strncpy(r,str+i,l1);
              s[l1]=r[l1]='\0';
              cmp(r);
              f=1;
            for(j=0;j<n;j++)
                if(strstr(a[j],s)==NULL && strstr(a[j],r)==NULL)
                {
                    f=0;
                    break;
                }
                if(f==1)
                    return l1;
          }
          l1--;
      }
      return 0;
}

int main()
{
    int i,minstrlen,substrlen,t;//minstrlen记录最短字符串的长度，substrlen返回最长子串长度
      char minstr[101];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        minstrlen=100;//先假设字符串最小长度值为100
        for(i=0;i<n;i++)
        {
               scanf("%s",a[i]);
             if(strlen(a[i])<minstrlen)//寻找n个字符串中最短的字符串
             {
                 strcpy(minstr,a[i]);
                 minstrlen=strlen(minstr);
             }
        }
        substrlen=search(minstr);
      printf("%d\n",substrlen);
    }
    return 0;
}