杭电-1238Substrings（爆搜）

Substrings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8626 Accepted Submission(s): 4001

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
There should be one line per test case containing the length of the largest string found.

[align=left]Sample Input[/align]

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

[align=left]Sample Output[/align]

2
2

猛地一看还以为是LCS，原来是个爆搜题！不会超时放心枚举！

找出最短的串，然后枚举这个串的所有子串和反串即可！

#include<cstdio>
#include<cstring>
#include<aLgorithm>
#define INF 0x3f3f3f3f
char map[110][110],s1[110],s2[110];
int main()
{
int N,k,i,j,m,n,ma,f,MAX;
scanf("%d",&N);
while(N--)
{
scanf("%d",&n);
ma=INF;
for(i=0;i<n;i++)
{
scanf("%s",map[i]);
int len=strlen(map[i]);
if(len<ma)
{
ma=len;
f=i;
}
}
MAX=0;
int flag=1;
for(i=0;i<ma;i++)
{
for(j=i;j<ma;j++)
{
for(k=i;k<=j;k++)
{
s1[k-i]=map[f][k];//正子串
s2[j-k]=map[f][k];//反子串
}
s1[j-i+1]=s2[j-i+1]='\0';

int len3=strlen(s1);
for(k=0;k<n;k++)
{
if(!strstr(map[k],s1)&&!strstr(map[k],s2))//子串都没出现过
{
flag=0;
break;
}
}
if(len3>MAX&&flag)
MAX=len3;
flag=1;
}
}
printf("%d\n",MAX);
}
return 0;
}