Mobile Service

这是一道很DT的DP，你认为呢？

const
	infinite=1000000000;
var
	w:array[0..200,0..200]of longint;
	n,i,j,x1,x2,ans,l:longint;
	p:array[0..200]of longint;
	f:array[0..3,0..200,0..200]of longint;
function min(a,b:longint):longint;
begin
	if a=-1 then exit(b) else
	if a<b then exit(a) else exit(b);
end;

begin
	fillchar(ans,sizeof(ans),$3f);
	filldword(f,sizeof(f)div 4,infinite);
	readln(l,n);
	for i:=1 to l do
	for j:=1 to l do
		read(w[i,j]);
	for i:=1 to n do
		read(p[i]);
	f[1,1,2]:=0;
	p[0]:=3;
	for i:=1 to n do
	begin
		for x1:=1 to l do
		for x2:=1 to l do
		begin
			f[i mod 2+1,p[i-1],x2]:=infinite;
			f[i mod 2+1,x1,p[i-1]]:=infinite;
			f[i mod 2 +1,x1,x2]:=infinite;
		end;
		for x1:=1 to l do
		for x2:=1 to l do
		if f[(i-1)mod 2+1,x1,x2]<>infinite then
		begin
			f[i mod 2+1,p[i-1],x2]:=min(f[i mod 2+1,p[i-1],x2],f[(i-1)mod 2+1,x1,x2]+w[x1,p[i]]);
			f[i mod 2+1,x1,p[i-1]]:=min(f[i mod 2+1,x1,p[i-1]],f[(i-1)mod 2+1,x1,x2]+w[x2,p[i]]);
			f[i mod 2 +1,x1,x2]:=min(f[i mod 2 +1,x1,x2],f[(i-1)mod 2+1,x1,x2]+w[p[i-1],p[i]]);
		end;
	end;
	for x1:=1 to l do
	for x2:=1 to l do
	begin
	
		if(f[n mod 2+1,x1,x2]<>-1)and(f[n mod 2+1,x1,x2]<ans) then ans:=f[n mod 2+1,x1,x2];
	end;
	writeln(ans);
end.	

对于这个滚存数组，是否有更好的写法呢？