HDU 1560 DNA sequence （迭代加深搜索）

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. 

InputThe 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.OutputFor 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
如果DNA最长的串长度为n,那就先搜索以n为长度，是否存在符合条件的母串，若不存在，再搜索n+1；
这便是所谓的迭代加深搜索。结束。
代码中用到的maxx，只是一个剪枝而已。

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1e9+7;
const double eps = 1e-6;

char s[12][12];
int maxs=0;
int tot[12];
int n,pos[12];
char a[5]="ACGT";

void view()
{
    for(int i=1;i<=n;i++){
        cout<<pos[i]<<" ";
    }
    cout<<endl;
}

bool dfs(int t)
{

    int maxx=0;
    for(int i=1;i<=n;i++){
        maxx=max(maxx,tot[i]-pos[i]);
    }
    if(maxx==0){return true;}
    if(t+maxx>maxs){return false;}
    bool vis[12];
    for(int i=0;i<4;i++){
        memset(vis,0,sizeof(vis));
        for(int j=1;j<=n;j++){
            if(s[j][pos[j]]==a[i]){
                pos[j]++;
                vis[j]=true;
            }
        }
        if(dfs(t+1)){return true;}
        for(int j=1;j<=n;j++){
            if(vis[j]){
                pos[j]--;
            }
        }
    }
    return false;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        maxs=0;
        for(int i=1;i<=n;i++){
            scanf("%s",s[i]);
            tot[i]=strlen(s[i]);
            maxs=max(maxs,tot[i]);
        }

        while(true){
            memset(pos,0,sizeof(pos));
            if(dfs(0)){
                printf("%d\n",maxs);
                break;
            }
            else maxs++;
        }
    }
    return 0;
}