HDU 1560 迭代加深 解题报告

DNA sequence

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

1
4
ACGT
ATGC
CGTT
CAGT

Sample Output

8

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))char str[10][10];int t,n,deep;int ans;char DNA[4]={'A','T','C','G'}; void dfs(int index,int len[]){    if(index>deep) return;            int maxx=0;              for(int i=0;i<n;++i)        maxx=max(strlen(str[i])-len[i],maxx);      if(maxx==0)    {           ans=index;          return;      }      if(index+maxx>deep) return;                 for(int i=0;i<4;++i)    {          int flag=0;          int pos[10];          for(int j=0;j<n;++j)        {              if(str[j][len[j]]==DNA[i])            {                flag=1;                  pos[j]=len[j]+1;              }              else pos[j]=len[j];          }          if(flag) dfs(index+1,pos);          if(ans!=-1) return;     }   }int main(){    for(scanf("%d",&t);t;--t)    {        deep=0;        scanf("%d",&n);        for(int i=0;i<n;++i)        {            scanf("%s",str[i]);            deep=max(deep,strlen(str[i]));        }        ans=-1;          int pos[10]={0};            while(1)        {              dfs(0,pos);              if(ans!=-1) break;              ++deep; //加深迭代          }          printf("%d\n",ans);      }}