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

第一次寫迭代深搜的題

參照了別人的題解 

搜尋層數是在主函式增加的 這玩意不用完整展開解答樹省時間

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define clr(a,b) memset(a,b,sizeof(a));
using namespace std;

bool flag=0;
char map[10][10],str[]="ACGT";
int point[10];//配對成功數量 
int num[10];//每行個數 
int n;

int estimation() //最差估價 
{
	int ans=0;
	for(int i=0;i<n;i++)
	{
		ans=max(ans,num[i]-point[i]);//一層搜可以匹配一個位置字元 如果最大為倆字元 搜兩層就夠了 
	}
	return ans;
}

void dfs(int maxstep)
{
	if(estimation()==0)//估價為0 搜到辣 
	{
		flag=true;
		return;
	}
	if(maxstep<estimation())//估價大於最大搜索層數 在主函式裡增加搜尋層數 
	{
		return;
	}
	int repetition[10];//儲存現場 
	for(int i=0;i<n;i++)
	repetition[i]=point[i];
	int yes=0;
	for(int i=0;i<4;i++)
	{
		for(int j=0;j<n;j++)
		{
			if(map[j][point[j]]==str[i])
			{
				yes=1;//本層有匹配成功的 減一層maxstep搜搜看 這樣就可以不用完整展開解答樹 
				point[j]++;
			}
		}	
		if(yes)		
		{
			dfs(maxstep-1);
			if(flag)
			return;
			for(int k=0;k<n;k++)
			point[k]=repetition[k];
		}			
		
	}
}

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		cin>>n;
		clr(point,0)
		clr(num,0)
		clr(map,'\0')
		flag=0;
		for(int i=0;i<n;i++)
		{
			scanf("%s",map[i]);
			num[i]=strlen(map[i]);
		}
		int result=0;
		for(int i=0;i<n;i++)
		result=max(result,num[i]);
		while(1)
		{
			dfs(result);
			if(flag==true)
			{
				cout<<result<<endl;
				break;
			}
			result++;
		}
	}
	return 0;	
}