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 换成' '; 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' '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]=' '; 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 }