HDU1560(KB2-E IDA星)

DNA sequence

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2427    Accepted Submission(s): 1198

Problem Description

The twenty-first century is a biology-technology developing century. We know that a gene is made of DNA. The nucleotide 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 be not the only one.

 

Input

The 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.

 

Output

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

 

Sample Input

1

4

ACGT

ATGC

CGTT

CAGT

 

Sample Output

8

 

Author

LL

对公共串的每一位进行搜索，并使用IDA*剪枝

 //2017-03-07
 #include <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 string DNA[];
 char ch[] = {'A', 'T', 'C', 'G'};
 int deep, n, pos[];//pos[i]表示第i个DNA串匹配到了pos[i]位
 bool ok;

 int Astar()//估价函数，返回当前状态最少还需要多少位
 {
     int h = ;
     for(int i = ; i < n; i++)
           if(h < DNA[i].length()-pos[i])
               h = DNA[i].length()-pos[i];
     return h;
 }

 void IDAstar(int step)//枚举所求串每一位的字符，step表示所求串当前处理到哪一位。
 {
     if(ok)return;
     int h = Astar();
     if(step + h > deep)return;//剪枝
     if(h == ){//h为0表示短串全部处理完
         ok = true;
         return;
     }
     for(int i = ; i < ; i++)
     {
         int tmp[];
         for(int j = ; j < n; j++)tmp[j] = pos[j];
         bool fg = false;
         for(int j = ; j < n; j++)
         {
             if(DNA[j][pos[j]] == ch[i]){
                 fg = true;//表示所求串第step为可以是ch[i]
                 pos[j]++;
             }
         }
         if(fg){
             IDAstar(step+);
         }
         for(int j = ; j < n; j++)pos[j] = tmp[j];
     }
 }

 int main()
 {
     int T;
     cin>>T;
     while(T--)
     {
         cin>>n;
         deep = ;
         for(int i = ; i < n; i++)
         {
             cin>>DNA[i];
             if(DNA[i].length() > deep)deep = DNA[i].length();
         }
         ok = false;
         while(!ok)
         {
             memset(pos, , sizeof(pos));
             IDAstar();
             deep++;
         }
         cout<<deep-<<endl;
     }

     return ;
 }