hdu 1560 DNA sequence （IDA*）

DNA sequence

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 721    Accepted Submission(s): 349

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
题意：很容易看懂
思路：dfs枚举所有情况，由于深度未知，所以可以利用迭代加深搜索的方式。
感想：开始用朴素dfs算法  果断超时  于是想到IDA*   由于一个数组定义为全局变量了  debug我好久   
代码：
#include <iostream>#include <cstdio>#include <cstring>#define maxn 10using namespace std;int n,ans,depth;int len[maxn],p[maxn];char s[maxn][maxn];char dx[]={'A','C','G','T'};int geth()                          // A*函数  得到最小需要几个字符来满足条件{    int i,j,t=-1,tt;    for(i=1;i<=n;i++)    {        tt=len[i]-p[i];        if(t<tt) t=tt;    }    return t;}bool dfs(int pos){    int i,j,h,flag;    int tmp[maxn];                   // 开始把这个定义为全局变量了 debug我好久 ╮(╯▽╰)╭    h=geth();                        // 得到最小需要几个字符来满足条件    if(pos+h>depth) return false ;   // 如果加上最小所需字符都不能满足条件则剪枝    if(h==0)    {        ans=pos;        return true ;    }    for(i=1;i<=n;i++)        tmp[i]=p[i];    for(i=0;i<4;i++)    {        flag=0;                         // 用来剪枝        for(j=1;j<=n;j++)        {            if(s[j][p[j]]==dx[i])            {                flag=1;                p[j]++;            }        }        if(!flag) continue ;            // 如果p[j]都没变化  则加的字母没用  剪枝        if(dfs(pos+1)) return true ;        for(j=1;j<=n;j++)           p[j]=tmp[j];    }    return false ;}int main(){    int i,j,t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        depth=-1;        for(i=1;i<=n;i++)        {            scanf("%s",s[i]);            len[i]=strlen(s[i]);            p[i]=0;            if(depth<len[i]) depth=len[i];        }        while(!dfs(0)) depth++;            // IDA*  每次限定深度        printf("%d\n",ans);    }    return 0;}