POJ 3080(最长公共子串）

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Blue Jeans

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8767 Accepted: 3688

Description

给出若干个基因串（由'A','T','S','C'构成），请找出最长公共子串。

Input

第一行为数据数。

对每组数据：第一行为字符串的个数m(2 <= m <= 10)

                        之后m行，每行一个基因串（有且仅有60个字母）

Output

对每组数据，找出最长公共子串，如果长度小于3，请输出 "no significant commonalities" ，否则输出最长的字符串，若有多个答案，输出字典序最小的。

Sample Input

32GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA3GATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATACCAGATAGATACTAGATACTAGATACTAGATACTAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAAGATACCAGATACCAGATACCAGATACCAAAGGAAAGGGAAAAGGGGAAAAAGGGGGAAAA3CATCATCATCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCACATCATCATAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAACATCATCATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

Sample Output

no significant commonalitiesAGATACCATCATCAT

Source

South Central USA 2006

枚举字符串更新答案，用kmp算法进行匹配。

Program BlueJeans;const   maxn=60;   maxm=10;var   tt,i,j,k,n,m:longint;   a:array[1..maxm] of string;   next:array[1..maxn] of longint;   p,ans:string;   flag:boolean;function kmp(a,s:string):boolean;var   i,j,n:longint;begin   n:=length(a);   j:=0; next[1]:=0; i:=1;   while i<n do   begin      if (j=0) or (a[i]=a[j]) then      begin         inc(i); inc(j);         if (a[i]<>a[j]) then next[i]:=j else next[i]:=next[j];      end else j:=next[j];   end;   j:=0; i:=0;   while (i<=maxn) and (j<=n) do   begin      if (j=0) or (a[j]=s[i]) then      begin         inc(i); inc(j);  //       if (j=n) then exit(true);      end else j:=next[j];   end;   if j>n then exit(true);   exit(false);end;function compare(a,b:string):boolean;var   i,n,m:longint;begin   n:=length(a); m:=length(b);   if (n<>m) then exit(n<m);   for i:=1 to n do if (a[i]<>b[i]) then exit(ord(a[i])>ord(b[i]));   exit(false);end;begin   readln(tt);   while (tt>0) do   begin      ans:='';      readln(m);      for i:=1 to m do readln(a[i]);      for i:=1 to maxn do         for j:=i to maxn do         begin            p:=copy(a[1],i,j-i+1);            flag:=false;            for k:=2 to m do            begin               if not(kmp(p,a[k])) then begin flag:=true; break; end;            end;            if not(flag) and (compare(ans,p)) then ans:=p;         end;      if length(ans)<3 then writeln('no significant commonalities')      else writeln(ans);      dec(tt);   end;end.