poj 3080 Blue Jeans

最新推荐文章于 2020-03-26 11:59:37 发布

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最新推荐文章于 2020-03-26 11:59:37 发布

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分类专栏： ACM算法_KMP 文章标签： KMP算法

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/78513700

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本文介绍了一个基于KMP算法的问题解决方案，旨在找出多个DNA序列中的最长公共子序列，用于基因研究中的模式识别。

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Blue Jeans

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19129		Accepted: 8511

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

题意就是求n个长度为60的串中求最长公共子序列（长度>=3）；如果有多个输出字典序最小的；

我们可以暴力求出第一个串的所有子串，然后判断是否是其他的子串即可；

KMP+暴力

从第一个串取出所有字串，去进行匹配。

#include <iostream>
#include <cstdio>
#include <cstring>
const int N = 107;
char str[N][65];
int Next[N];
using namespace std;
void getNext(char a[], int n)
{
    Next[0] = -1;
    int i = 0, j = -1;
    while(i < n)
    {
        if(j == -1 || a[i] == a[j])
        {
            i++;
            j++;
            Next[i] = j;
        }
        else
            j = Next[j];
    }
}
int kmp(char a[], char b[], int n)
{
    int i = 0, j = 0;
    while(i < 60)
    {
        if(j == -1 || b[i] == a[j])
        {
            i++;
            j++;
        }
        else
            j = Next[j];
        if(j == n)
            return true;
    }
    return false;
}
int main()
{
    int T, f;
    scanf("%d", &T);
    while(T--)
    {
        int m;
        scanf("%d", &m);
        for(int i = 0; i < m; i++)
            scanf("%s", str[i]);
        char ans[N] = "a";
        for(int i = 60; i >= 3; i--) //字符串长度
        {
            for(int j = 0; j <= 60 - i; j++) //起始位置
            {
                char subStr[N] = {0};
                strncpy(subStr, str[0] + j, i);
                getNext(subStr, i);
                int flag = 0;
                for(int k = 1; k < m; k++)
                {
                    if(!kmp(subStr, str[k], i)) 
                    {
                        flag = 1;
                        break;
                    }
                }
                if(!flag && strcmp(ans, subStr) > 0)
                {
                    strcpy(ans, subStr);
                }
            }
            f=0;
            if (ans[0]!='a')
            {
            	f=1;
            	printf("%s\n",ans);
            	break;
            }
        }
        if(!f)
        	printf("no significant commonalities\n");
    }
    return 0;
}

参考网址

https://www.cnblogs.com/zhengguiping--9876/p/4838631.html