Hdu1560 DNA sequence（IDA*） 2017-01-20 18:53 50人阅读 评论(0) 收藏

DNA sequence

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

Total Submission(s) : 15   Accepted Submission(s) : 7

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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

————————————————————————————————————————————————

题意：从ｎ个串中找出一个最短的公共串（也许应该说序列吧，因为不要求连续，即只要保持相对顺序就好）。



分析：迭代加深搜索，就是每次都限制了ＤＦＳ的深度，若搜不到答案，则加深深度，继续搜索，这样就防止了随着深度不断加深而进行的盲目搜索，而且，对于这种求最短长度之类的题目，只要找到可行解，即是最优解了。所以就这样敲完代码了，敲完之后，悲剧ＴＬＥ。



少了一步十分重要的剪枝，就是每次ＤＦＳ的时候，都要判断一下，当前的深度＋最少还有加深的深度是否大于限制的长度，若是，则退回。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
char s[10][10];
int n,mx,flag;
char ch[5]={'A','T','C','G'};

void dfs(int deep,int *cnt)
{
    if(flag)
        return;
    int sum=0;
    int maxlen=0;
    int a;
    for(int i=0; i<n; i++)
    {
        a=strlen(s[i])-cnt[i];
        sum=sum+a;
        maxlen=max(maxlen,a);
    }
    if(maxlen+deep>mx)
        return;
    if(sum==0)
    {
        flag=1;
        return;
    }
    int fl=0;
    int next[10];
    for(int i=0; i<4; i++)
    {
        for(int j=0; j<n; j++)
        {
            if(s[j][cnt[j]]==ch[i])
        {
            fl=1;
            next[j]=cnt[j]+1;
            }
            else
                next[j]=cnt[j];
            }
        if(fl)
        {
            dfs(deep+1,next);
        }
        if(flag)
            return;
    }

}

int main()
{
    int o,k;
    int cnt[10];
    scanf("%d",&o);
    while(o--)
   {
    scanf("%d",&n);
        mx=0;
        for(int i=0; i<n; i++)
        {
            scanf("%s",s[i]);
            k=strlen(s[i]);
            mx=max(mx,k);
        }
        memset(cnt,0,sizeof(cnt));
        flag=0;
        for(int i=0; i<40; i++)
        {
            dfs(0,cnt);

            if(flag)
                break;
                mx++;
        }
        printf("%d\n",mx);
    }
    return 0;
}