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[0..105] of string;
    next:array[0..105] of integer;
    f:array[0..105] 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),0);
    i:=1;
    j:=0;
    next[1]:=0;
    while i<=l do
    begin
      if (j=0) 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:=1;
    j:=1;
    l1:=length(a);
    while (i<=l1) and (j<=l) do
    begin
      if (j=0) 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:=1 to n do
    begin
      if (j=v) or f[j] then continue;
      if (compare(a[j],s)) then
      begin
        f[j]:=true;
        w:=w+1;
      end;
    end;
  end;

begin
  readln(t);
  for p:=1 to t do
  begin
    readln(n);
    v:=0;
    for i:=1 to n do
    begin
      readln(a[i]);
      if (v=0) or (length(a[i])<length(a[v])) then v:=i;
    end;
    for l:=length(a[v]) downto 1 do
    begin
      for i:=1 to length(a[v])-l+1 do
      begin
        w:=1;
        fillchar(f,sizeof(f),false);
        s:=copy(a[v],i,l);
        work_next(s);
        kmp;
        for j:=1 to l div 2 do
        begin
          c:=s[j];
          s[j]:=s[l-j+1];
          s[l-j+1]:=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(0);
  end;
end.