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POJ 3080, Blue Jeans
阅读量:7155 次
发布时间:2019-06-29

本文共 3121 字,大约阅读时间需要 10 分钟。

brute force

用第一个字符串从长到短生成所有的子串, 并和其他字符串匹配。

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

South Central USA 2006


//
 POJ3080.cpp : Defines the entry point for the console application.
//
#include 
<
iostream
>
#include 
<
string
>
using
 
namespace
 std;
string
 ComSubStr(
string
*
 ln, 
int
 n)
{
    
string
 longest
=
""
;
    
for
 (
int
 j 
=
 
60
; j 
>
 
2
--
j)
        
for
 (
int
 i 
=
 
0
; i 
+
 j 
<=
 
60
++
i)
        {
            
string
 str 
=
 ln[
0
].substr(i, j);
            
bool
 find 
=
 
true
;
            
for
 (
int
 k 
=
 
1
; k 
<
 n; 
++
k)
                
if
 (ln[k].find(str) 
==
 
string
::npos) 
                {
                    find 
=
 
false
;
                    
break
;
                };
            
            
if
 (find 
==
 
true
 
&&
 str.size() 
>
 longest.size())longest 
=
 str;
            
else
 
if
(find 
==
 
true
 
&&
 str.size() 
==
 longest.size() 
&&
 str 
<
 longest) longest 
=
 str;
        };
    
return
 longest;
};
int
 main(
int
 argc, 
char
*
 argv[])
{
    
int
 cases;
    cin 
>>
 cases;
    
int
 lines;
    
string
 ln[
10
];
    
for
 (
int
 c 
=
 
0
; c 
<
 cases; 
++
c)
    {
        cin 
>>
 lines;
        
for
 (
int
 i 
=
 
0
; i 
<
 lines; 
++
i)
        {
            cin 
>>
 ln[i];
        }
        
        
string
 r 
=
 ComSubStr(ln, lines);
        
if
 (r 
==
 
""
) cout 
<<
 
"
no significant commonalities\n
"
;
        
else
 cout 
<<
 r 
<<
 endl;
    }
    
return
 
0
;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/01/1577214.html

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