poj3080 Blue Jeans(串的暴力枚举)

最新推荐文章于 2019-01-14 17:08:23 发布

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本文介绍了一种用于比对多个DNA序列并找出最长公共子序列的算法实现。该算法适用于基因组研究，如《国家地理》与IBM合作的《基因图谱计划》，通过分析DNA贡献者的数据来绘制地球人口分布图。文章提供了一个具体的示例代码，展示了如何通过逐个比较序列来确定共同的遗传标记。

Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16401		Accepted: 7283

Description

The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated.

As an IBM researcher, you have been tasked with writing a program that will find commonalities amongst given snippets of DNA that can be correlated with individual survey information to identify new genetic markers.

A DNA base sequence is noted by listing the nitrogen bases in the order in which they are found in the molecule. There are four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). A 6-base DNA sequence could be represented as TAGACC.

Given a set of DNA base sequences, determine the longest series of bases that occurs in all of the sequences.

Input

Input to this problem will begin with a line containing a single integer n indicating the number of datasets. Each dataset consists of the following components:

A single positive integer m (2 <= m <= 10) indicating the number of base sequences in this dataset.
m lines each containing a single base sequence consisting of 60 bases.

Output

For each dataset in the input, output the longest base subsequence common to all of the given base sequences. If the longest common subsequence is less than three bases in length, display the string "no significant commonalities" instead. If multiple subsequences of the same longest length exist, output only the subsequence that comes first in alphabetical order.

Sample Input

3
2
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
3
GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
3
CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalities
AGATAC
CATCATCAT

Source

意：给你几个DNA序列长度为60，以第一个为模板，找到之后的DNA中与模板DNA相同的子序列，且保证子序列最长（长度大于等于3）。

依旧是暴力找就行，枚举相同序列的长度以第一个DNA为模板向其他串中找。其中有个技巧性的地方就是strstr（）函数的使用，

#include<stdio.h>
#include<string.h>
char str[100][61];
char tem[61];
char fin[61];
int n;

int judge()
{
   for(int i =1 ;i<n;i++)
   {
       if(strstr(str[i],tem)==0)//strstr(a,b)函数为在a中找b，如果可以找到b那么会返回最初始找到b时的位置的地址，若找不到b则返回NULL。
           return 0;
    }
    return 1;
}

int main()
{
    int t;
    int m,i,j;
    scanf("%d",&t);
    while(t--)
    {
        memset(fin,0,sizeof(fin));
        memset(str,0,sizeof(str));
        scanf("%d",&n);

        for(int i = 0; i<n; i++)
            scanf("%s",str[i]);


        int k = 3;
        int flag=0;

        while(k<=60)//枚举字母的个数
        {
            memset(tem,0,sizeof(tem));
            for(int i = 0; i <=60 - k; i++)//枚举主串的起点
            {
                for(int j=0;j<k;j++)
                    tem[j]=str[0][i+j];//储存在临时的数组中
                     tem[k]='\0';

              if(judge())//判断枚举的串是否是其余串的子串
              {
                  flag=1;
                  memcpy(fin,tem,sizeof(tem));//
              }

            }
            k++;
        }
        if(flag==1)
            printf("%s\n",fin);
        else
         printf("no significant commonalities\n");
    }

    return 0;

}