题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1560

DNA sequence

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

Total Submission(s): 1505 Accepted Submission(s): 730

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

Author

Source

HDU 2006-12 Programming Contest

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题目大意：给n个序列，找到一个包含所有给出序列的最短长度并输出。

解题思路：采用gei_h()得到当前状态下最长的未匹配的长度。在进行深度搜索。每个串的长度不超过5，最多只有8个序列，所以IDA不超过40次。

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>using namespace std;int n;
char ch[10][10];
int len[10],want;
char dir[10]= {'A','C','G','T'};
int wei[10];//记录第i个序列正在使用第几个位置int get_h()
{int t=0;for (int i=1; i<=n; i++){t=max(t,len[i]-wei[i]);//得到当前情况下最长的未被匹配的长度}return t;
}int IDA(int dep)
{if(dep+get_h()>want)//当前长度+估测的长度比我想要的还大的话，就不必继续搜索{//cout<<get_h()<<endl;return 0;}if(dep==want)return 1;int tmp[10];for (int i=0; i<4; i++){int flag=0;memcpy(tmp,wei,sizeof(wei));//先存一下for (int j=1; j<=n; j++){if (ch[j][wei[j]]==dir[i])//当前的这一位符合{flag=1;//标记当前状态可以wei[j]++;}}if (flag){if(IDA(dep+1))//如果可以就继续搜索return 1;memcpy(wei,tmp,sizeof(tmp));//还原回来}}return 0;
}int main()
{int T;scanf("%d",&T);while (T--){int Max=0;scanf("%d",&n);for (int i=1; i<=n; i++){scanf("%s",ch[i]);len[i]=strlen(ch[i]);if (len[i]>Max)Max=len[i];}memset(wei,0,sizeof(wei));want=Max;//从最长序列开始查找while (1){if (IDA(0)){break;}want++;}printf ("%d\n",want);}return 0;
}

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