POJ-3080-Blue Jeans
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Blue Jeans
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16962 Accepted: 7526
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
32GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA3GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAGATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAAGATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA3CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCACATCATCATcuchunAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
Sample Output
no significant commonalitiesAGATACCATCATCAT
Source
South Central USA 2006
题目的意思大概就是求多组字符串的最长公共子序列,并且长度小于3的输出“no significant commonalities“;
暴力枚举,啦啦啦!!!
一直都听人说暴力,那就暴力一发,毕竟MAX是60位字符;
#include <iostream>#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int len = 60;
char Map[11][61];//储存所有字符
int main()
{
int t, m, i, j;
scanf("%d", &t);
while (t--)
{
scanf("%d", &m);
memset(Map, 0, sizeof(Map));
for (i = 0; i < m; i++)
{
scanf("%s", Map[i]);//记录所有的字符串
}
int length = 1;
char Maxchar[61];//公共串
int Maxlength = 0; //最大串的长度
for (i = 0;; i++)
{
char nowchar[61];
int pi = i;
if (pi + length > len)
{
length++;
i = -1;
if (length > len)
break;
continue;
}
for (j = 0; j < length; j++)
{
nowchar[j] = Map[0][pi++];
}
nowchar[j] = '\0';
bool flag = true;
for (int l = 1; l < m; l++)
{
if (!strstr(Map[l], nowchar))
{
flag = false;
break;
}
}
if (flag)
{
if (Maxlength < length)
{
Maxlength = length;
strcpy(Maxchar, nowchar);
}
else if (Maxlength == length)
{
if (strcmp(Maxchar, nowchar) > 0)
strcpy(Maxchar, nowchar);
}
}
}
if (Maxlength < 3)
printf("no significant commonalities\n");
else
printf("%s\n", Maxchar);
}
return 0;
}
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