题目链接:http://poj.org/problem?id=3080
Blue Jeans
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 16228 | Accepted: 7220 |
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.
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
题目大意:给你n串长度为60的字符串,问他们的最长公共子串为什么,长度小于三的输出那串英文,大于三输出公共子串。这个,直接暴力就好,因为长度为60,所以怎么暴力怎么来
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char str[15][65];
char sub[65];
char ans[65];
int len;
int main()
{
int t,i,j,k,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(ans,'\0',sizeof(ans));
for(i = 1; i<=n; i++)
scanf("%s",str[i]);
for(i = 1; i<=60; i++)
{
int find = 0;
for(j = 0; j<=60-i; j++)
{
len = 0;
int check = 1;
for(k = j;;k++)
{
sub[len++] = str[1][k];
if(len == i)
break;
}
sub[len] = '\0';
for(k = 2; k<=n; k++)
{
if(!strstr(str[k],sub))
{
check = 0;
break;
}
}
if(check)
{
find = 1;
if(strlen(ans)<strlen(sub))
strcpy(ans,sub);
else if(strcmp(ans,sub)>0)
strcpy(ans,sub);
}
}
if(!find)
break;
}
if(strlen(ans)<3)
printf("no significant commonalities\n");
else
printf("%s\n",ans);
}
return 0;
}