kuangbin专题十六 KMP&&扩展KMP POJ3080 Blue Jeans
发布时间:2020-12-14 05:14:37 所属栏目:大数据 来源:网络整理
导读: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 taske
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.
Input
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 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:
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 AACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTSample Output no significant commonalities AGATAC CATCATCAT 1 #include<iostream> 2 #include<stdio.h> 3 #include<string> 4 #include<set> 5 #include<vector> 6 using namespace std; 7 vector<string> t; 8 set<string> ss; 9 string s; 10 int _,n; 11 12 string fun() { 13 ss.clear(); 14 string str=t[0]; 15 bool flag; 16 for(int len=60;len>=3;len--) { 17 for(int ix=0;ix<=60-len;ix++) { 18 string temp=str.substr(ix,len); 19 flag=true; 20 for(int k=1;k<t.size();k++) { 21 if(t[k].find(temp)==-1) { 22 flag=false; 23 break; 24 } 25 } 26 if(flag) ss.insert(temp); 27 } 28 if(ss.size()) return *ss.begin(); 29 } 30 return "no significant commonalities"; 31 } 32 33 int main() { 34 // freopen("in","r",stdin); 35 for(scanf("%d",&_);_;_--) { 36 scanf("%d",&n); 37 for(int i=0;i<n;i++) { 38 cin>>s; 39 t.push_back(s); 40 } 41 cout<<fun()<<endl; 42 t.clear(); 43 } 44 45 } ? kmp思想:不需要找第一个串的所有子串,只需枚举每一个后缀,去和其他字符串匹配就行了。其实这个匹配过程就好比所有子串进行匹配了。 1 #include<stdio.h> 2 #include<iostream> 3 #include<string> 4 #include<algorithm> 5 #include<vector> 6 using namespace std; 7 int _,n,Next[61]; 8 string s,strans; 9 vector<string> t; 10 11 void prekmp(string s) { 12 int len=s.size(); 13 int i,j; 14 j=Next[0]=-1; 15 i=0; 16 while(i<len) { 17 while(j!=-1&&s[i]!=s[j]) j=Next[j]; 18 if(s[++i]==s[++j]) Next[i]=Next[j]; 19 else Next[i]=j; 20 } 21 } 22 23 int kmp(string p,string t) { 24 int len=t.size(); 25 int i=0,j=0,res=-1; 26 while(i<len) { 27 while(j!=-1&&t[i]!=p[j]) j=Next[j]; 28 ++i;++j; 29 res=max(res,j); 30 } 31 return res; 32 } 33 34 35 int main() { 36 // freopen("in",stdin); 37 for(scanf("%d",&_);_;_--) { 38 scanf("%d",&n); 39 for(int i=0;i<n;i++) { 40 cin>>s; 41 t.push_back(s); 42 } 43 int ans=-1; 44 string str=t[0]; 45 for(int i=0;i<60;i++) { 46 string temp=str.substr(i,60-i); 47 prekmp(temp); 48 int maxx=60; 49 for(int j=1;j<t.size();j++) { 50 maxx=min(maxx,kmp(temp,t[j])); 51 } 52 if(maxx>ans) { 53 strans=temp.substr(0,maxx); 54 ans=maxx; 55 } else if(maxx==ans) { 56 string anstemp=temp.substr(0,maxx); 57 if(anstemp<strans) strans=anstemp; 58 } 59 } 60 if(strans.size()<3) cout<<"no significant commonalities"<<‘n‘; 61 else cout<<strans<<‘n‘; 62 t.clear(); 63 } 64 } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |