Evolution

Time Limit: 20000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Every kind of living creatures has a kind of DNA. The nucleotide bases from which DNA is built are A (adenine), C (cytosine), G (guanine), and T (thymine). Sometimes if two DNA of two living creatures have the same substring, and
the length is beyond a certain percentage of the whole length, we many consider whether the two living creatures have the same ancestor. And we can separate them into a certain species temporarily for our research, and we say the two living creatures are similar
Make sure if A is similar with B, and B is similar with C, but C is not similar with A, we also separate A, B and C into a kind, for during the evolution, there happens aberrance.

Now we have some kinds of living creatures and their DNA, just tell us how many kinds of living creatures we can separate.

Input

There are a lot of cases. In each case, in the first line there are two numbers N and P. N means the number of kinds of living creatures. If two DNA are similar, there exist a substring, and its length is beyond the percentage of
any DNA of the two, and P is just the percentage. And 1<=N<=100, and 1<=P<100 (P is 100, which means two DNA are similar if and only if they are the same, so we make sure P is smaller than 100). The length of each DNA won‘t exceed 100.

Output

For each case, just print how many kinds living creatures we can separate.

Sample Input

3 10.0
AAA
AA
CCC

Sample Output

Case 1:
2

#include<stdio.h>
#include<string.h>
const int N = 105;

int fath[N],ins[N],n;

void init()
{
    for(int i=0;i<n;i++)
        fath[i]=i,ins[i]=0;
}
int findfath(int x)
{
    if(x!=fath[x])
        fath[x]=findfath(fath[x]);
    return fath[x];
}
void setfath(int x,int y)
{
    x=findfath(x);
    y=findfath(y);
    fath[x]=y;
}
int main()
{
    double P;
    int c=0,len[N];
    char DNA[N][N];
    while(scanf("%d%lf",&n,&P)>0)
    {
        for(int i=0;i<n;i++)
        {
             scanf("%s",DNA[i]);
             len[i]=strlen(DNA[i]);
        }

        init();
        for(int i=0; i<n; i++)
        for(int j=i+1; j<n; j++)
        {
           int maxlen=0,flag=0;
            for(int ti=0;ti<len[i]&&maxlen<len[i]-ti;ti++)
            {
                for(int tj=0;tj<len[j]&&maxlen<len[j]-tj;tj++)
                {
                    int ii,jj;
                    for( ii=ti,jj=tj; ii<len[i]&&jj<len[j]; ii++,jj++)
                    if(DNA[i][ii]!=DNA[j][jj])
                    {
                       break;
                    }
                    if(ii-ti>maxlen)
                        maxlen=ii-ti;
                    if(100*maxlen/(len[i]*1.0)>P&&100*(maxlen/(len[j]*1.0))>P)
                            setfath(i,j),flag=1;
                    if(flag)break;
                }
                if(flag)break;
            }
        }
        int k=0;
        for(int i=0;i<n;i++)
            if(fath[i]==i)
            k++;
        printf("Case %d:\n%d\n",++c,k);
    }
}