poj1226,poj3080

看来以后用pascal的函数要小心了；

简简单单pos其实时间复杂度是二次方级的……

今天学习的是KMP——字符匹配算法；

这两道题也都很简单，都是为这个算法练手的，

最朴素的匹配显然是穷举起始位置然后看是否匹配，复杂度O(nm)不尽人意

kmp的思想就是尽可能利用之前匹配的信息进行匹配。

具体分析我就不讲了，传送门http://www.cppblog.com/oosky/archive/2006/07/06/9486.html讲的很详细

但据说这两题暴力都可过……

 var a:array[..] of string;
     next:array[..] of integer;
     f:array[..] of boolean;
     s:string;
     c:char;
     i,j,l,t,p,v,n,w:integer;
 procedure work_next(s:string);
   var i:integer;
   begin
     fillchar(next,sizeof(next),);
     i:=;
     j:=;
     next[]:=;
     while i<=l do
     begin
       if (j=) or (s[i]=s[j]) then
       begin
         inc(i);
         inc(j);
         next[i]:=j;
       end
       else j:=next[j];
     end;
   end;
 function compare(a,b:string):boolean;
   var i,j,l1:integer;
   begin
     i:=;
     j:=;
     l1:=length(a);
     while (i<=l1) and (j<=l) do
     begin
       if (j=) or (a[i]=b[j]) then
       begin
         inc(i);
         inc(j);
       end
       else j:=next[j];
     end;
     if j>l then exit(true) else exit(false);
   end;

 procedure kmp;
   var j:integer;
   begin
     for j:= to n do
     begin
       if (j=v) or f[j] then continue;
       if (compare(a[j],s)) then
       begin
         f[j]:=true;
         w:=w+;
       end;
     end;
   end;

 begin
   readln(t);
   for p:= to t do
   begin
     readln(n);
     v:=;
     for i:= to n do
     begin
       readln(a[i]);
       if (v=) or (length(a[i])<length(a[v])) then v:=i;
     end;
     for l:=length(a[v]) downto  do
     begin
       for i:= to length(a[v])-l+ do
       begin
         w:=;
         fillchar(f,sizeof(f),false);
         s:=copy(a[v],i,l);
         work_next(s);
         kmp;
         for j:= to l div  do
         begin
           c:=s[j];
           s[j]:=s[l-j+];
           s[l-j+]:=c;
         end;
         work_next(s);
         kmp;
         if w=n then break;
       end;
       if w=n then break;
     end;
     if w=n then writeln(l) else writeln();
   end;
 end.

poj1226