bzoj3294

感觉自己就是不怎么擅长计数的问题

设f[k,i,j]表示前k种颜色占据了i行j列的方案

g[k,i,j]表示第k种颜色占据了i行j列的方案，注意要减去并没占据满i行j列的情况

然后转移就很好写了

像这种题目构造好了状态后都非常好解决

const mo=1000000009;
var g,f:array[0..11,0..31,0..31] of longint;
    c:array[0..1010,0..1010] of longint;
    a:array[0..15] of longint;
    n,m,i,j,k,q,x,y,ans:longint;

begin
  readln(n,m,q);
  c[0,0]:=1;
  for i:=1 to n*m do
  begin
    c[i,0]:=1;
    for j:=1 to i do
      c[i,j]:=(c[i-1,j]+c[i-1,j-1]) mod mo;
  end;
  for i:=1 to q do
    read(a[i]);
  for k:=1 to q do
    for i:=1 to n do
      for j:=1 to m do
      begin
        if (i*j<a[k]) or (i>a[k]) or (j>a[k]) then continue;
        g[k,i,j]:=c[i*j,a[k]];
        for x:=1 to i do
          for y:=1 to j do
            if (i-x+j-y)>0 then
              g[k,i,j]:=(g[k,i,j]-int64(g[k,x,y])*int64(c[i,x]) mod mo*int64(c[j,y]) mod mo+mo) mod mo;
      end;

  f[0,0,0]:=1;
  for k:=1 to q do
  begin
    a[k]:=a[k]+a[k-1];
    for i:=1 to n do
      for j:=1 to m do
      begin
        if i*j<a[k] then continue;
        for x:=1 to i do
          for y:=1 to j do
            f[k,i,j]:=(f[k,i,j]+int64(f[k-1,i-x,j-y])*int64(g[k,x,y]) mod mo*int64(c[i,x]) mod mo*int64(c[j,y]) mod mo) mod mo;
      end;
  end;
  for i:=1 to n do
    for j:=1 to m do
      ans:=(ans+int64(f[q,i,j])*int64(c[n,i]) mod mo*int64(c[m,j]) mod mo) mod mo;

  writeln(ans);
end.