POJ-3080 Blue Jeans（纯暴力）

C - Blue Jeans

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice POJ 3080

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

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这道题但是真的想了太久，一直纳闷kmp怎么跑n个字符，

后来查了题解才知道原来无脑暴力就能过，真是醉了。

大致题意和思路：

找出n个串中最长的公共串，并且要求字典序最大

直接枚举第一串的所有子串，然后与后面的所有串进行比较即可

拓展：<p style="margin-top: 0px; margin-bottom: 0px; padding-top: 0px; padding-bottom: 0px; color: rgb(69, 69, 69); font-family: arial, 宋体, sans-serif, tahoma, 'Microsoft YaHei'; font-size: 14px; line-height: 24px;">strstr(str1,str2)；</p><p style="margin-top: 0px; margin-bottom: 0px; padding-top: 0px; padding-bottom: 0px; color: rgb(69, 69, 69); font-family: arial, 宋体, sans-serif, tahoma, 'Microsoft YaHei'; font-size: 14px; line-height: 24px;"> 函数用于判断字符串str2是否是str1的子串。如果是，则该函数返回str2在str1中首次出现的地址；否则，返回NULL。</p><p style="margin-top: 0px; margin-bottom: 0px; padding-top: 0px; padding-bottom: 0px; color: rgb(69, 69, 69); font-family: arial, 宋体, sans-serif, tahoma, 'Microsoft YaHei'; font-size: 14px; line-height: 24px;"><pre name="code" class="cpp">#include<stdio.h>#include<string.h>#include<stdlib.h>#include<limits.h>#include<algorithm>using namespace std;int n,m;char a[15][65];char b[65],c[65];int  main(){int T,i,j;scanf("%d",&T);while(T--){memset(c,'\0',sizeof(c));scanf("%d",&n);for(i=1;i<=n;i++)scanf("%s",a[i]);for(i=1;i<=60;i++){int flag=0;for(j=0;j<=60-i;j++){int len=0;int check=1;for(int k=j;;k++){b[len++]=a[1][k];if(len==i)break;}b[len]='\0';for(int k=2;k<=n;k++){if(!strstr(a[k],b)){check=0;break;}}if(check){flag=1;if(strlen(c)<strlen(b))strcpy(c,b);if(strcmp(c,b)>0)strcpy(c,b);}}if(!flag)break;}if(strlen(c)<3)printf("no significant commonalities\n");elseprintf("%s\n",c);}}