HDU 1560 DNA sequence (IDA *)

DNA sequence

Time Limit: 15000/5000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)

Total submission (s): 721 accepted submission (s): 349 problem descriptionthe twenty-first century is a biology-technology developing century. we know that a gene is made of DNA. the nucleus otide bases from which DNA is built are a (adenine), C (cytosine), g (guanine), and T (thymine ). finding the longest common
Subsequence between DNA/protein sequences is one of the basic problems in modern computational molecular biology. but this problem is a little different. given several DNA sequences, you are asked to make a shortest sequence from them so that each of the given
Sequence is the subsequence of it.

For example, given "acgt", "atgc", "cgtt" and "cagt", you can make a sequence in the following way. it is the shortest but may not the only one.

 

Inputthe first line is the test case number t. then T test cases follow. in each case, the first line is an integer N (1 <= n <= 8) represents number of the DNA sequences. the following K lines contain the K sequences, one per line. assuming
That the length of any sequence is between 1 and 5.

Outputfor each test case, print a line containing the length of the shortest sequence that can be made from these sequences.

Sample Input

14ACGTATGCCGTTCAGT

 

Sample output

8. Meaning: it is easy to understand the concept: DFS enumerates all cases. Because the depth is unknown, you can use iterations to deepen the search. Feeling: I started to use the simple DFS algorithm to time out, so I thought of IDA * because an array is defined as a global variable, I 've been coding for a long time:
# Include <iostream> # include <cstdio> # include <cstring> # define maxn 10 using namespace STD; int N, ANS, depth; int Len [maxn], P [maxn]; char s [maxn] [maxn]; char DX [] = {'A', 'C', 'G','t '}; int geth () // A * function requires a minimum of several characters to meet the conditions {int I, j, t =-1, TT; for (I = 1; I <= N; I ++) {TT = Len [I]-P [I]; If (T <TT) t = tt;} return t ;} bool DFS (int pos) {int I, j, H, flag; int TMP [maxn]; // start to define this as a global variable. Debug: I have been debugging for a long time. Then, when response H = geth (); // it takes a few characters to get the minimum value. If (Pos + H> depth) return false; // if the minimum required character cannot meet the condition, the request is trimmed if (H = 0) {ans = Pos; return true;} for (I = 1; I <= N; I ++) TMP [I] = P [I]; for (I = 0; I <4; I ++) {flag = 0; // used for pruning (j = 1; j <= N; j ++) {If (s [J] [p [J] = DX [I]) {flag = 1; p [J] ++ ;}} if (! Flag) continue; // If P [J] does not change, the added letter will not use pruning if (DFS (Pos + 1) return true; For (j = 1; j <= N; j ++) P [J] = TMP [J];} return false;} int main () {int I, j, T; scanf ("% d", & T); While (t --) {scanf ("% d", & N); depth =-1; for (I = 1; I <= N; I ++) {scanf ("% s", s [I]); Len [I] = strlen (s [I]); P [I] = 0; If (depth <Len [I]) depth = Len [I];} while (! DFS (0) Depth ++; // ida * specifies the depth of printf ("% d \ n", ANS) each time;} return 0 ;}