poj1088

这题是dp还是dfs+记忆化？（其实好像没什么区别？）

用f[i,j]表示滑到(i,j)时之后最多能滑多远，依次穷举每一个起点(i,j)则

f[i,j]=max{f[i,j-1],f[i-1,j],f[i+1,j],f[i,j+1]}+1 （当然滑到那个点的高度要小于(i,j)的高度）

 const dx:array[..] of integer=(,,-,);
       dy:array[..] of integer=(-,,,);
 var f,a:array[..,..] of longint;
     n,i,j,m,ans:longint;
 function max(a,b:longint):longint;
   begin
     if a>b then max:=a else max:=b;
   end;
 procedure search(x,y:longint);
   var i,xx,yy:integer;
       maxx:longint;
   begin
     maxx:=;
     for i:= to  do
     begin
       xx:=x+dx[i];
       yy:=y+dy[i];
       if (a[xx,yy]<a[x,y]) and (f[xx,yy]=) then search(xx,yy);
       if (a[xx,yy]<a[x,y]) then maxx:=max(maxx,f[xx,yy]);
     end;
     f[x,y]:=maxx+;
   end;

 begin
   readln(n,m);
   for i:= to n+ do
     for j:= to m+ do
       a[i,j]:=;
   for i:= to n do
   begin
     for j:= to m do
       read(a[i,j]);
     readln;
   end;
   for i:= to n  do
     for j:= to m do
       if f[i,j]= then search(i,j);
   for i:= to n do
     for j:= to m do
       ans:=max(ans,f[i,j]);
   writeln(ans);
 end.