POJ308--Blue Jeans--后缀树

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

32GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA3GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAGATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAAGATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA3CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalitiesAGATACCATCATCAT
#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;#define maxn 20008int pre[maxn];struct suffixtree{int ch[maxn][26];int val[maxn];int sz;suffixtree(){sz=1;memset(ch,0,sizeof(ch));}int idx(char c){return c-'A';}void insert(char *s,int v){int len=strlen(s);int u=0;for(int i=0;i<len;i++){int c=idx(s[i]);if(!ch[u][c]){ch[u][c]=sz++;}pre[ch[u][c]]=u;u=ch[u][c];if(val[u]==v-1){val[u]=v;}}}}wa;int main(){int t;scanf("%d",&t);while(t--){int maxlen=0;for(int i=1;i<maxn;i++){pre[i]=i;}memset(wa.ch,0,sizeof(wa.ch));memset(wa.val,0,sizeof(wa.val));int n;scanf("%d",&n);char A[68];for(int i=1;i<=n;i++){cin>>A;wa.insert(A,i);char B[68];for(int j=1;j<60;j++){int m=0;for(int k=j;k<60;k++){B[m++]=A[k];}B[m]='\0';wa.insert(B,i);}}//然后扫描一遍所有节点，如果节点的val为n。就一直找他的父亲节点。找一段最长的val都为n的节点段char C[68],D[68];for(int i=wa.sz;i>=0;i--){if(wa.val[i]==n){int maxlen_=0;int j=i;while(wa.val[j]==n){maxlen_++;if(j==pre[j]||j==0){break;}j=pre[j];}int m=i;for(int k=maxlen_-1;k>=0;k--){int z=pre[m];for(int l=0;l<26;l++){if(wa.ch[z][l]==m){D[k]='A'+l;break;}}m=pre[m];}if(maxlen_>maxlen){maxlen=maxlen_;for(int j=0;j<maxlen;j++){C[j]=D[j];}C[maxlen]='\0';}else if(maxlen_==maxlen&&strcmp(C,D)==1){maxlen=maxlen_;for(int j=0;j<maxlen;j++){C[j]=D[j];}C[maxlen]='\0';}}}if(maxlen<3){cout<<"no significant commonalities"<<endl;}else cout<<C<<endl;}return 0;}