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

 1 //寻找最长字串并输出其长度
 2 //  思路：
 3 //  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
 4 //  2、根据str搜索满足条件的子字符串
 5 // 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
 6 //strlen：计算字符串长度
 7 //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。
 8 //strncpy：复制字符串中的字串，它的用法要掌握。
 9 //strcpy：复制字符串
10 //怎样找最短字符串，也是值得学习的。。
11 //搜索满足条件的最长子串.
12 #include<stdio.h>
13 #include<string.h>
14 char a[101][101];
15 int n;
16 void cmp(char *str)//字符串反序
17 {
18     int l,i,g=0;
19     char b[101];
20     l=strlen(str);
21     for(i=0;i<l;i++)
22         b[g++]=str[l-i-1];
23     b[g]='\0';
24     strcpy(str,b);
25 }
26 int search(char *str)//搜索满足条件的最长子串并返回其长度
27 {
28     int  l1,l2,i,j,f;
29     char s[101],r[101];
30     l1=strlen(str);l2=strlen(str);
31       while(l1>0)//搜索不同长度的子串，从最长的子串开始搜索
32       {
33           for(i=0;i<=l2-l1;i++)
34           {
35               strncpy(s,str+i,l1);//strncpy的函数用法必须是str+i不能写成str【i】；
36               strncpy(r,str+i,l1);
37               s[l1]=r[l1]='\0';
38               cmp(r);
39               f=1;
40             for(j=0;j<n;j++)
41                 if(strstr(a[j],s)==NULL && strstr(a[j],r)==NULL)
42                 {
43                     f=0;
44                     break;
45                 }
46                 if(f==1)
47                     return l1;
48           }
49           l1--;
50       }
51       return 0;
52 }
53 
54 int main()
55 {
56     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
57       char minstr[101];
58     scanf("%d",&t);
59     while(t--)
60     {
61         scanf("%d",&n);
62         minstrlen=100;//先假设字符串最小长度值为100
63         for(i=0;i<n;i++)
64         {
65                scanf("%s",a[i]);
66              if(strlen(a[i])<minstrlen)//寻找n个字符串中最短的字符串
67              {
68                  strcpy(minstr,a[i]);
69                  minstrlen=strlen(minstr);
70              }
71         }
72         substrlen=search(minstr);
73       printf("%d\n",substrlen);
74     }
75     return 0;
76 }