HDU ACM 1238 Substrings

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6004 Accepted Submission(s): 2657

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

ABCD

BCDFF

BRCD

rose

orchid

Sample Output

Author

Asia 2002, Tehran (Iran), Preliminary

Recommend

Eddy

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define SIZE 102
 7 
 8 using namespace std;
 9 
10 typedef struct strType
11 {
12     char value[SIZE];
13     int len;
14 }strType;
15 
16 strType alpha[SIZE];
17 char pos[SIZE], neg[SIZE];
18 
19 bool Traverse(int cur, int neglect, int n)
20 {
21     if(cur == neglect) return Traverse(cur+1, neglect, n);
22     if(cur == n) return true;
23     
24     if(strstr(alpha[cur].value, pos) != NULL || strstr(alpha[cur].value, neg) != NULL)
25         return Traverse(cur+1, neglect, n);
26     else return false;
27 }
28 
29 int main()
30 {
31     int t;
32     scanf("%d", &t);
33     while(t--)
34     {
35         int n, shortest, id;
36         scanf("%d", &n);
37         getchar();
38         for(int i=0; i<n; ++i)
39         {
40             scanf("%s", alpha[i].value);
41             alpha[i].len = strlen(alpha[i].value);
42             if(!i) shortest = alpha[i].len, id = i;
43             else if(shortest > alpha[i].len)
44             {
45                 shortest = alpha[i].len;
46                 id = i;
47             }
48         }
49         int res = 0;
50         for(int i=alpha[id].len-1; i>=0; i--)
51         {
52             if(i+1 < res) break;
53             for(int j=0; j<=i; j++)
54             {
55                 if(i-j+1 < res) break;
56                 memcpy(pos, alpha[id].value+j, sizeof(char)*(i-j+1));
57                 neg[i-j+1] = pos[i-j+1] = '\0';
58                 for(int t=0; t<i-j+1; ++t) neg[i-j-t] = pos[t];
59                 if(Traverse(0, id, n))
60                 {
61                     res = i-j+1;
62                     if(res < i-j+1) res = i-j+1;
63                 }
64             }
65         }
66         printf("%d\n", res);    
67     }
68     return 0;
69 }