poj1038

【题意】
有个N*M个格子芯片,其上有k个坏点，问放2*3的组件最多能放多少个
N (1 <= N <= 150), M (1 <= M <= 10)
【输入】
第一行是个D，表示多少组数据
第二N，M，K
然后K行描述哪几个点坏了
【输出】
每组数据输出一个数，表示最多能放多少个。

M很小，所以每行可以状压表示
状态压缩压两行

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.