[BZOJ1087][SCOI2005]互不侵犯King解题报告|状压DP

在N×N的棋盘里面放K个国王，使他们互不攻击，共有多少种摆放方案。国王能攻击到它上下左右，以及左上左下右上右下八个方向上附近的各一个格子，共8个格子。

　　好像若干月前非常Naive地去写过DFS...

　　然后其实作为状压DP是一道非常好的题啦><

　　感觉直接无脑搞时间是下不来的 做了好几道预处理

　　使得最后DP的过程中没有任何一条转移是无用的

　　

 program bzoj1087;
 var i,x,n,k,j,p,q,t1,t2:longint;
     ans:int64;
     a:array[-..]of longint;
     b:array[-..,-..]of longint;
     c:array[-..,-..]of longint;
     f:array[..,-..,-..]of int64;

 function check(x:longint):boolean;
 var las:longint;
 begin
     las:=;
     while x<> do
     begin
         if (x and =)and(las=) then exit(false);
         las:=x and ;
         x:=x >> ;
     end;
     exit(true);
 end;

 function count(x:longint):longint;
 var tem:longint;
 begin
     tem:=;
     while x<> do
     begin
         inc(tem,x and );
         x:=x >> ;
     end;
     exit(tem);
 end;

 function ok(x,y:longint):boolean;
 var i:longint;
     tmp:array[..,-..]of longint;
 begin
     for i:= downto  do tmp[,i]:=x >> i and ;
     for i:= downto  do tmp[,i]:=y >> i and ;
     for i:= to  do if (tmp[,i]+tmp[,i]=)or(tmp[,i]+tmp[,i-]=)or(tmp[,i]+tmp[,i+]=) then exit(false);
     exit(true);
 end;

 begin
     fillchar(b,sizeof(b),);
     fillchar(a,sizeof(a),);
     fillchar(c,sizeof(c),);
         readln(n,k);
         for i:= to  << n- do if check(i) then
     begin
             x:=count(i);
         inc(b[x,]);b[x,b[x,]]:=i;
         inc(a[]);a[a[]]:=i;
     end;
     for i:= to a[] do
         for j:= to a[] do if ok(a[i],a[j]) then begin inc(c[a[i],]);c[a[i],c[a[i],]]:=a[j];end;
     fillchar(f,sizeof(f),);
     for i:= to k do
         for j:= to b[i,] do f[,i,b[i,j]]:=;
     for i:= to n do
     begin
         f[]:=f[];
         fillchar(f[],sizeof(f[]),);
         for j:= to k do
             for p:= to (n+) >>  do if p<=j then
             begin
                 q:=j-p;
                 for t1:= to b[p,] do
                     for t2:= to c[b[p,t1],] do inc(f[,j,b[p,t1]],f[,q,c[b[p,t1],t2]]);
             end;
     end;
     ans:=;
     for i:= to a[] do inc(ans,f[,k,a[i]]);
     writeln(ans);
 end.