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  • POJ3080Blue Jeans

    转载请注明出处：優YoU http://user.qzone.qq.com/289065406/blog/1309012790

     

    大致题意：

    就是求k个长度为60的字符串的最长连续公共子串，2<=k<=10

    规定：

    1、  最长公共串长度小于3不输出

    2、  若出现等长的最长的子串，则输出字典序最小的串

     

    解题思路：

    纠结了几个月放着没做的题目。。

    一直以为要用KMP或者后缀数组来做。。。

    然后我就拼命学后缀。。。

    今天偶然发现直接 暴力 能够达到0ms的效果= =

    所以。。。暴力吧。。。不愧为初级的题。。。

     

    暴力思想很简单：

    开二维DNA[][]保存所有DNA序列

     

    1、  以DNA[0]为母版，顺次截取60个长度length=1的子串dna[]，检查其他DNA[i]是否都有子串dna，若是则把dna[]复制到obj[]，否则枚举下一个长度length的子串；若obj与新的dna等长，则比较两者字典序，当dna字典序小时，复制到obj

    2、  第1步循环60次后length+1。

    顺次截取59个长度length=2的子串dna[]，重复1的操作更新obj。。

    3、  第2步循环59次后length+1。

    顺次截取58个长度length=3的子串dna[]，继续。。

    ...........

    60、第59步循环2次后length+1。

    顺次截取1个长度length=60的子串dna[]，继续重复操作更新obj。。

    61、输出obj

     

    Source修正：

    http://acm2006.cct.lsu.edu/problems/

     

     

     1 //Memory Time 
     2 //248K   16MS 
     3 
     4 #include<iostream>
     5 #include<string.h>
     6 using namespace std;
     7 
     8 const int len=60;
     9 
    10 int main(int i,int j)
    11 {
    12     int test;
    13     cin>>test;
    14     for(int t=1;t<=test;t++)
    15     {
    16         int n;  //DNA个数
    17         cin>>n;
    18         char** DNA=new char*[n];
    19         for(int p=0;p<n;p++)
    20         {
    21             DNA[p]=new char[len+1];
    22             cin>>DNA[p];
    23         }
    24 
    25         char obj[len+1];  //所有DNA的公共串
    26         int StrLen=0;    //最长公共串长度
    27         int length=1;    //当前枚举的公共串长度
    28 
    29         for(i=0;;i++)  //枚举公共串dna[]
    30         {
    31             /*截取DNA[0][]中以pi为起点，长度为length的子串dna[]*/
    32             char dna[len+1];
    33             int pi=i;
    34             if(pi+length>len)
    35             {
    36                 length++;
    37                 i=-1;
    38                 if(length>len)
    39                     break;
    40                 continue;
    41             }
    42             for(j=0;j<length;j++)
    43                 dna[j]=DNA[0][pi++];
    44             dna[j]='\0';
    45 
    46             /*检查其他DNA[][]是否都存在字符串dna[]*/
    47             bool flag=true;
    48             for(int k=1;k<n;k++)
    49                 if(!strstr(DNA[k],dna))  //存在一个DNA不含有dna[]
    50                 {
    51                     flag=false;
    52                     break;
    53                 }
    54 
    55             /*确认最大公共串*/
    56             if(flag)
    57             {
    58                 if(StrLen<length)
    59                 {
    60                     StrLen=length;
    61                     strcpy(obj,dna);    
    62                 }
    63                 else if(StrLen==length)
    64                 {
    65                     if(strcmp(obj,dna)>0)  //存在相同长度的公共串时，取最小字典序的串
    66                         strcpy(obj,dna);
    67                 }
    68             }
    69         }
    70 
    71         if(StrLen<3)
    72             cout<<"no significant commonalities"<<endl;
    73         else
    74             cout<<obj<<endl;
    75 
    76         delete DNA;
    77     }
    78     return 0;
    79 }

    Sample Input
     
    11
    2
    GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    3
    GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATA
    GATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
    GATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA
    3
    CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
    ACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
    2
    GATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGAT
    GATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATT
    10
    GATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGATGAT
    GATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATTGATT
    GATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTTGATTT
    GATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTTGATTTT
    AGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGAT
    AAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGATAAGAT
    AAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGATAAAGAT
    CGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGAT
    CCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGATCCGAT
    CCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGATCCCGAT
    10
    GATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATC
    GCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTA
    TAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGCTAGC
    CGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGATCGAT
    GTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGTGT
    GAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGAGA
    GCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGCGC
    ATATATATATATATATATATATATATATATATATATATATATATATATATATATATATAT
    ACACACACACACACACACACACACACACACACACACACACACACACACACACACACACAC
    TCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTCTC
    10
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    10
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAG
    GGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAG
    GGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGG
    GGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGG
    GGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGG
    GGGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGG
    GGGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGGG
    GGGGGAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAGGGG
    4
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG
    TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
    CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
    10
    AAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAAAATAAAA
    AAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAACAAAAAC
    AAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAGAAAAAG
    AAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAATAAAAAT
    AAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAACAAAAC
    AAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAGAAAAG
    AAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAATAAAAT
    AAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAACAAAC
    AAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAATAAAT
    AAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAG
    2
    GATGATGCATCATGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGACTACTAA
    GATGATCATCATACTACTCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

     

    Sample Output
    no significant commonalities
    AGATAC
    CATCATCAT
    TGAT
    GAT
    no significant commonalities
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
    no significant commonalities
    AAA
    ACTACT

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  • 原文地址：https://www.cnblogs.com/lyy289065406/p/2122174.html
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