(KMP 暴力)Corporate Identity -- hdu -- 2328
http://acm.hdu.edu.cn/showproblem.php?pid=2328
Corporate Identity
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 698 Accepted Submission(s): 281
After several other proposals, it was decided to take all existing trademarks and find the longest common sequence of letters that is contained in all of them. This sequence will be graphically emphasized to form a new logo. Then, the old trademarks may still be used while showing the new identity.
Your task is to find such a sequence.
After the last trademark, the next task begins. The last task is followed by a line containing zero.
IDENTITY LOST
代码:
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std; #define M 40005
#define N 210 char s[M][N];
int Next[N]; void FindNext(char b[])
{
int i=, j=-, blen=strlen(b);
Next[] = -; while(i<blen)
{
if(j==- || b[i]==b[j])
Next[++i] = ++j;
else
j = Next[j];
}
} int KMP(char a[], char b[])
{
int i=, j=;
int alen=strlen(a), blen=strlen(b); FindNext(b); while(i<alen)
{
while(j==- || (a[i]==b[j] && i<alen && j<blen))
i++, j++;
if(j==blen)
return ;
j = Next[j];
}
return ;
} int main()
{
int n;
while(scanf("%d", &n), n)
{
int i, j, k, MinLen=, len;
char ss[N]; memset(s, , sizeof(s)); for(i=; i<n; i++)
{
scanf("%s", s[i]);
len = strlen(s[i]); if(len<MinLen)
{
MinLen = len;
memset(ss, , sizeof(ss));
strcpy(ss, s[i]);
}
} char b[N]="{";
int index=; for(i=MinLen; i>; i--)
{
for(j=; j<=MinLen-i; j++)
{
char a[N]; memset(a, , sizeof(a)); strncpy(a, ss+j, i); for(k=; k<n; k++)
{
if(KMP(s[k], a)==)
break;
} if(k==n && strcmp(a, b)<)
{
index = i;
strcpy(b, a);
} if(index && j==MinLen-i)
i=-, j=;
}
} if(index)
printf("%s\n", b);
else
printf("IDENTITY LOST\n"); }
return ;
}
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