HDU 1560DNA sequence

DNA sequence

Time Limit : 15000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 1

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

14ACGTATGCCGTTCAGT

 

Sample Output

8

 

Author

LL

 

Source

HDU 2006-12 Programming Contest

找到最短的DNA序列  其子序列包含题意所给的DNA

IDA*  估价函数为最长剩余DNA数量

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<cmath>#include<string>#include<vector>#include<algorithm>#include<map>#include<set>#define inf 1<<30#define LL long long#define maxn 1<<24using namespace std;char str[10][10];//保存DNAint l[10];int t,n;int tlen;bool flag;int get_h(int * a)  //估价函数{    int ans=0;    for(int i=0; i<n; i++)    {        ans=max(ans,a[i]);    }    return ans;}void dfs(int len,int * a){    if(flag) return ;    if(get_h(a)>len) return ;//预估值小于最优估计值    if(len==0)  //找到序列    {        flag=true ;        return  ;    }    bool vis[10];    memset(vis,false ,sizeof(vis));    for(int i=0; i<n; i++)    {        if(a[i]==0||vis[i]) continue ;        int ee[10];        for(int ii=0; ii<n; ii++)            ee[ii]=a[ii];        vis[i]=true ;        char ch=str[i][l[i]-a[i]];        ee[i]--;        for(int j=i+1; j<n; j++)        {            if(ee[j]==0||vis[j]) continue ;            if(str[j][l[j]-a[j]]==ch)            {                vis[j]=true ;                ee[j]--;            }        }        dfs(len-1,ee);    }    return  ;}int main(){    scanf("%d",&t);    while(t--)    {        tlen=0;        flag=false ;        scanf("%d",&n);        int ee[10];        for(int i=0; i<n; i++)        {            scanf("%s",str[i]);            ee[i]=strlen(str[i]);            l[i]=ee[i];            tlen=max(ee[i],tlen);        }        while(1)        {            dfs(tlen,ee);            if(flag)                break;            tlen++;        }        printf("%d\n",tlen);    }}