poj3167

这道题是一道kmp的扩展版的好题
一串匹配一串很容易想到kmp，但是这里的匹配要求的是两个串的名次相同
显然名次是会变的，为了方便，我们可以换一种表达
对于两个等长的串的相同位置，名次相等就是在它之前比它小的数的个数一样，和它相等的数的个数一样
这个我们可以用树状数组维护一下（当然暴力好像也行）
然后匹配就行了

 var ans,a,b,sal,eq:array[..] of longint;
     next:array[..] of longint;
     c:array[..] of longint;
     tot,k,n,m,s,i,j:longint;

 function lowbit(x:longint):longint;
   begin
     exit(x and (-x));
   end;

 procedure work(x,p:longint);
   begin
     while x<=s do
     begin
       inc(c[x],p);
       x:=x+lowbit(x);
     end;
   end;

 function ask(x:longint):longint;
   begin
     ask:=;
     while x> do
     begin
       ask:=ask+c[x];
       x:=x-lowbit(x);
     end;
   end;

 begin
   readln(n,m,s);
   for i:= to n do
     readln(a[i]);
   for i:= to m do   //预处理
   begin
     readln(b[i]);
     work(b[i],);
     sal[i]:=ask(b[i]-);
     eq[i]:=ask(b[i]);
   end;
   fillchar(c,sizeof(c),);
   i:=;
   j:=;
   next[]:=;
   while i<=m do
   begin
     if (j=) or (ask(b[i]-)=sal[j]) and (ask(b[i])=eq[j]) then
     begin
       inc(i);
       inc(j);
       next[i]:=j;
       if i>m then break;
       work(b[i],);
     end
     else begin
       for k:=i-j+ to i-next[j] do  // 保证匹配的串位置一样
         work(b[k],-);
       j:=next[j];
     end;
   end;
   fillchar(c,sizeof(c),);
   i:=;
   j:=;
   work(a[],);
   while (i<=n) do
   begin
     if (j=) or (ask(a[i]-)=sal[j]) and (ask(a[i])=eq[j]) then
     begin
       inc(i);
       inc(j);
       if i<=n then
         work(a[i],);
     end
     else begin
       for k:=i-j+ to i-next[j] do
         work(a[k],-);
       j:=next[j];
     end;
     if j>m then  //注意这里要找出的是所有答案
     begin
       for k:=i-m to i-next[j] do
         work(a[k],-);
       inc(tot);
       ans[tot]:=i-m;
       j:=next[j];
     end;
   end;
   writeln(tot);
   for i:= to tot do
     writeln(ans[i]);
 end.