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

這道題是對退出DFS的判斷問題，首先要判斷一下最長的有多長，然後當
這個節點對所有的DNA都沒有貢獻時就退出，這裡用到了迭代加深搜尋，就是我每一次搜尋都指定一個最大搜索長度，如果大於這個搜尋長度就退出，然後再深入一層，防止無限次的加深搜尋

#include<bits/stdc++.h>
using namespace std;
#define bug cout<<"This is a BUG!"<<endl;
using LL=int64_t;
const int INF=0x3f3f3f3f;
int n,deep,ans=-1;
char s[10][10];
int size[10];

char cnt[5]={'A','T','C','G'};

void dfs(int num,int len[]) {
    int sum=0;
    if(ans!=-1||num>deep) return;
    for(int i=0;i<n;i++) sum=max(sum,size[i]-len[i]);
    if(num+sum>deep) return ;
    if(sum==0) {ans=num;return ;}
    for(int i=0;i<4;i++) {
        int flag=0;
        int tlen[10]={0};
        for(int j=0;j<n;j++) {
            if(s[j][len[j]]==cnt[i]) {
                tlen[j]=len[j]+1;
                flag++;
            }
            else tlen[j]=len[j];
        }
        if(flag>0)
        dfs(num+1,tlen);
    }
    return ;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin>>T;
    for(int k=0;k<T;k++) {
        ans=-1,deep=0;
        cin>>n;
        memset(s,0,sizeof(s));
        for(int i=0;i<n;i++) {
            cin>>s[i];
            size[i]=strlen(s[i]);
            deep=max(deep,size[i]);
        }
        int len[10]={0};
        while(1) {
            dfs(0,len);
            if(ans!=-1) break;
            deep++;
        }
        cout<<ans<<endl;
    }
    return 0;
}