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=;
 var g,f:array[..,..,..] of longint;
     c:array[..,..] of longint;
     a:array[..] of longint;
     n,m,i,j,k,q,x,y,ans:longint;

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