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[10];
char ch[4] = {'A', 'T', 'C', 'G'};
int deep, n, pos[10];//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 = 0;
    for(int i = 0; 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 == 0){//h为0表示短串全部处理完
        ok = true;
        return;
    }
    for(int i = 0; i < 4; i++)
    {
        int tmp[10];
        for(int j = 0; j < n; j++)tmp[j] = pos[j];
        bool fg = false;
        for(int j = 0; 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+1);
        }
        for(int j = 0; j < n; j++)pos[j] = tmp[j];
    }
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        deep = 0;
        for(int i = 0; i < n; i++)
        {
            cin>>DNA[i];
            if(DNA[i].length() > deep)deep = DNA[i].length();
        }
        ok = false;
        while(!ok)
        {
            memset(pos, 0, sizeof(pos));
            IDAstar(0);
            deep++;
        }
        cout<<deep-1<<endl;
    }

    return 0;
}