学步园

3进制表示
0表示都没有
1表示第一行有
2表示第二行有
明白算法以后还是很好写的
=、=最大的错误是y+3写成了y+4
还有一个错误对于之前的不同状态递推过来的的dp要重新做一遍

program poj1038;
type
  bas=array [0..11] of longint;
var
  n,m,i,j,k,x,y,l,d,o,r,ans:longint;
  yes:array [0..151,0..11] of boolean;
  f:array [0..1,0..59049] of longint;
  three:array [0..11] of longint;
  q,p:bas;
function max (a,b:longint):longint;
begin
  if a>b then exit(a)
         else exit(b);
end;
function hash (q:bas):longint;
var
  i,ans:longint;
begin
  ans:=0;
  for i:=1 to m do
    ans:=ans+three[i-1]*q[i];
  exit(ans);
end;
function rehash (r:longint):bas;
var
  i:longint;
begin
  fillchar(rehash,sizeof(rehash),0);
  for i:=1 to m do
    begin
      rehash[i]:=r mod 3;
      r:=r div 3;
    end;
end;
procedure dfs (now,y,r:longint);
var
  i:longint;
begin
  if now>ans then ans:=now;
  if (y+1<=m)and(q[y]=0)and(q[y+1]=0)and(p[y]=0)and(p[y+1]=0) then
    begin
      p[y]:=2;
      p[y+1]:=2;
      i:=hash(p);
      if now+1>f[o,i] then f[o,i]:=now+1;
      dfs(now+1,y+2,i);
      p[y]:=0;
      p[y+1]:=0;
    end;
  if (y+2<=m)and(p[y]=0)and(p[y+1]=0)and(p[y+2]=0) then
    begin
      p[y]:=2;
      p[y+1]:=2;
      p[y+2]:=2;
      i:=hash(p);
      if now+1>f[o,i] then f[o,i]:=now+1;
      dfs(now+1,y+3,i);
      p[y]:=0;
      p[y+1]:=0;
      p[y+2]:=0;
    end;
  if now>f[o,r] then f[o,r]:=now;
  if (y+1<=m) then dfs(now,y+1,r);
end;
begin
  three[0]:=1;
  for i:=1 to 11 do
    three[i]:=three[i-1]*3;
  read(d);
  for l:=1 to d do
    begin
      fillchar(yes,sizeof(yes),false);
      read(n,m,k);
      for i:=1 to k do
        begin
          read(x,y);
          yes[x,y]:=true;
        end;
      for i:=1 to m do
        if yes[1,i] then q[i]:=2
                    else q[i]:=1;
      o:=0;
      fillchar(f,sizeof(f),255);
      r:=hash(q);
      f[o,r]:=0;
      ans:=0;
      for i:=2 to n do
        begin
          o:=1-o;
          fillchar(f[o],sizeof(f[o]),255);
          for j:=0 to three[m]-1 do
            if f[1-o,j]<>-1 then
              begin
                q:=rehash(j);
                for k:=1 to m do
                  if yes[i,k] then p[k]:=2
                              else
                  if q[k]<2 then p[k]:=0
                            else p[k]:=1;
                r:=hash(p);
                dfs(f[1-o,j],1,r);
              end;
        end;
      writeln(ans);
    end;
end.