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Substrings(hd1238)

时间:2015-06-23 15:01:42      阅读:142      评论:0      收藏:0      [点我收藏+]

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Substrings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8183    Accepted Submission(s): 3752


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
找到最短字符串,比如其长度为5,先取5的子串,再取4的子串...每次遍历其他字符串中是否含有其逆串或顺串
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 /* 原型:char * strncpy(char *dest, char *src, size_t n);   功能:将字符串src中最多n个字符复制到字符数组dest中(它并不像strcpy一样只有遇到NULL才停止复制,而是多了一个条件停止,就是说如果复制到第n个字符还未遇到NULL,也一样停止),返回指向dest的指针。*/
 6 int main()
 7 {
 8     int N,i,j,num;
 9     freopen("in.txt","r",stdin);
10     cin>>N;
11     while(N--)
12     {
13         char string[100][103],pos[103],inv[103],str[103];
14         int min_str=100,index,len,flag=0;
15         cin>>num;
16         for(i=0;i<num;i++)
17         {
18             cin>>string[i];//相当于把\n换成‘\0‘;
19             if(strlen(string[i])<min_str)
20                 min_str=strlen(string[i]);
21             index=i;
22         }//输入字符串,并找到短字符串
23         len=min_str;
24         strcpy(str,string[index]);
25         while(len>0)
26         {
27             for(i=0;i<=min_str-len;i++)//对于len长的子串可以取多少种;从最长的开始取
28             {
29                 flag=1;//假设该子串符合
30                 strncpy(pos,str+i,len);//并不会把‘\0‘复制进去,自己加进去
31                 for(j=0;j<len;j++)
32                     inv[j]=pos[len-j-1];//求出逆向子串;
33                 inv[len]=pos[len]=\0;
34                 for(j=0;j<num;j++)
35                 {
36                     if(strstr(string[j],inv)==NULL&&strstr(string[j],pos)==NULL)
37                     {
38                         flag=0;
39                         break;
40                     }
41                 }
42                 if(flag)
43                     break;
44             }
45             if(flag)
46                 break;
47             len--;
48         }
49         cout<<len<<endl;
50     }
51 }

 

Substrings(hd1238)

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原文地址:http://www.cnblogs.com/a1225234/p/4595290.html

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