poj3080 Blue Jeans(串的暴力枚举)

Blue Jeans

Time Limit: 1000MS Memory Limit: 65536KTotal 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

32GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA3GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAGATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAAGATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA3CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalitiesAGATACCATCATCAT

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;}