HDU 1560 - DNA sequence

E - DNA sequence

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1560

Appoint description: System Crawler  (2016-01-21)

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 

IDA*，temp[]记录每个字符串搜索到第几位

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <string>using namespace std;#define N 10struct node{    int len;    string s;}a[N];int n;int pos[N],depth,ans;char dna[]="ACGT";int calen(int *s){    int m=0;    for (int i=0;i<n;i++)        m=max(m,a[i].len-s[i]);    return m;}bool dfs(int *s,int now){    if (now+calen(s)>depth)        return false;    if (calen(s)==0)        return true;    int temp[N];    for (int i=0;i<4;i++){        bool flag=false;        memcpy(temp,s,sizeof(temp));        for (int j=0;j<n;j++){            if (a[j].s[temp[j]]==dna[i]){                flag=true;                temp[j]++;            }        }        if (flag){            if (dfs(temp,now+1))                return true;        }    }    return false;}int main(){    int t;    scanf("%d",&t);    while (t--){        scanf("%d",&n);        for (int i=0;i<n;i++){            cin>>a[i].s;            a[i].len=a[i].s.length();        }        memset(pos,0,sizeof(pos));        depth=1;        while (1){            if (dfs(pos,0))                break;            depth++;        }        printf("%d\n",depth);    }    return 0;}