POJ3080-Blue 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

posted on 2011-07-30 20:38  小優YoU  阅读(1256)  评论(0)  编辑  收藏  举报

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