Hdu2144evolution (and check set + violence or LCS)

EvolutionTime limit:20000/5000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 598 Accepted Submission (s): 143


Problem Description every kind of living creatures has a kind of DNA. The nucleotide bases from which DNA was built is A (adenine), C (cytosine), G (guanine), and T (thymine). Sometimes if the DNA of the living creatures has the same substring, and the length is beyond a certain percentage of the Whole length, we many consider whether the living creatures have the same ancestor. And we can separate them into a certain species temporarily for our we, and we say the both living creatures are simi Lar Make sure if A are similar with B, and B are similar with C, but C was not similar with A, we also separate A, B and C in To 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 is a lot of cases. In each case, the first line there is numbers n and P. N means the number of kinds of living creatures. If the DNA is similar, there exist a substring, and its length was beyond the percentage of any DNA of the the, and P is J UST the percentage. and 1<=n<=100, and 1<=p<100 (P is, which means-similar if and only if they was 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 10.0AAAAACCC

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]&&am                    p;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); }}

Hdu2144evolution (and check set + violence or LCS)