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
思路:
IDA*算法(迭代加深算法),详见代码注释
#include <iostream>
#include <string>
#include <cstring>
using namespace std;
int n;
char dna[4]={'A','T','C','G'};
string s[10];
int pos[10];///字串在所给串中匹配到的位置
int len[10];///字串的长度
int maxn;///所求串最短的长度
int geth()///h()函数,所求串最少还要在增长多少
{
int m=0;
for(int i=1;i<=n;i++)
m=max(m,len[i]-pos[i]);///len-pos得出字串未匹配的长度
return m;
}
bool solve(int dep)///dfs
{
if(dep+geth()>maxn) return false;
if(!geth()) return true;
int temp[10];
memcpy(temp,pos,sizeof(pos));///将pos拷贝到temp中,方便回溯
for(int i=0;i<4;i++)
{
int sign=0;
for(int j=1;j<=n;j++)
///答案串加入dna[i],与第j个字串的pos[j]匹配,则pos[j]++
if(s[j][pos[j]]==dna[i])
sign=1,pos[j]++;
if(sign)///答案串新加入的dna[i]有与字串匹配的,则往下一层搜索
{
if(solve(dep+1)) return true;///下一层符合,返回
memcpy(pos,temp,sizeof(pos));///回溯
}
}
return false;
}
int IDA_Star()///IDA*算法,迭代加深,直到满足题意
{
while(!solve(0)) maxn++;
return maxn;
}
int main()
{
int t;cin>>t;
while(t--)
{
memset(pos,0,sizeof(pos));
memset(len,0,sizeof(len));
cin>>n;
for(int i=1;i<=n;i++) cin>>s[i],len[i]=s[i].size();
maxn=0;
IDA_Star();
cout<<maxn<<endl;
}
return 0;
}