瓷砖铺放 （状压DP+矩阵快速幂）

由于方块最多涉及3行，于是考虑将每两行状压起来，dfs搜索每种状态之间的转移。

这样一共有2^12种状态，显然进行矩阵快速幂优化时会超时，便考虑减少状态。

进行两遍bfs，分别为初始状态可以到达的状态，和可以到达终止状态的状态。

同时出现在两次bfs中的状态即为有效状态，一共有141种。

这样就可以跑出来了。

未加矩阵快速幂 50分

 const dx:array[..,..] of longint=
     ((-,,),(-,,),(,,),(,,),(-,,),(-,,),(,,),(-,,));
     dy:array[..,..] of longint=
     ((,,),(,-,),(,-,),(,,),(,,-),(,,-),(,,-),(,,));
     mo=;
 var n,m,lx,i,j,k:longint;
     a:array[..,..] of longint;
     g:array[..,..] of longint;
     dp:array[..,..] of longint;
 function ok(x,y:longint):boolean; inline;
 begin
     if (x>=) and (x<=) and (y>=) and (y<=m) and (a[x,y]=) then exit(true);
     exit(false);
 end;
 function check(y,j:longint):boolean;
 var i,x:longint;
 begin
     x:=;
     if not ok(x,y) then exit(false);
     for i:= to  do
         if not ok(x+dx[j,i],y+dy[j,i]) then exit(false);
     exit(true);
 end;
 procedure fill(y,j,cl:longint);
 var i,x:longint;
 begin
     x:=;
     a[x,y]:=cl;
     for i:= to  do
         a[x+dx[j,i],y+dy[j,i]]:=cl;
 end;
 procedure dfs(x,lt:longint);
 var i,j,sum:longint;
 begin
     if x=m+ then
     begin
         for i:= to m do if a[,i]= then exit; //保证了DP的正确性
         sum:=;
         for i:= to  do
             for j:= to m do
                 sum:=sum*+a[i,j];
         inc(g[lt][sum]);
         exit;
     end;
     dfs(x+,lt);
     for i:= to  do
         if check(x,i) then
         begin
             fill(x,i,);
             dfs(x+,lt);
             fill(x,i,);
         end;
 end;
 begin
     assign(input,'tile.in');reset(input);
     assign(output,'tile.out');rewrite(output);
     readln(n,m);
     for i:= to <<(*m)- do
     begin
         for j:= to m do
             if i and (<<(m*-j))> then a[,j]:= else a[,j]:=;
         for j:= to m do
             if i and (<<(m-j))> then a[,j]:= else a[,j]:=;
         dfs(,i);
     end;
     dp[][(<<m-)<<m]:=;
     for i:= to n+ do
         for j:= to <<(*m)- do
             for k:= to <<(*m)- do
                 dp[i][j]:=(dp[i][j]+dp[i-][k]*g[k][j]) mod mo;
     writeln(dp[n+][(<<m-)<<m]);
     close(input);
     close(output);
 end.

AC 代码1：

 const dx:array[..,..] of longint=
     ((-,,),(-,,),(,,),(,,),(-,,),(-,,),(,,),(-,,));
     dy:array[..,..] of longint=
     ((,,),(,-,),(,-,),(,,),(,,-),(,,-),(,,-),(,,));
     mo=;
 type arr=array[..,..] of int64;
 var i,j,k,m,xh:longint;
     n:int64;
     a:array[..,..] of longint;
     g:array[..,..] of longint;
     b:array[..] of longint;
     v1,v2:array[..] of boolean;
     pos:array[..] of int64;
     mat,f:arr;
 function ok(x,y:longint):boolean; inline;
 begin
     if (y>=) and (y<=m) and (a[x,y]=) then exit(true);
     exit(false);
 end;
 function check(y,j:longint):boolean; inline;
 var i,x:longint;
 begin
     x:=;
     if not ok(x,y) then exit(false);
     for i:= to  do
         if not ok(x+dx[j,i],y+dy[j,i]) then exit(false);
     exit(true);
 end;
 procedure fill(y,j,cl:longint); inline;
 var i,x:longint;
 begin
     x:=;
     a[x,y]:=cl;
     for i:= to  do
         a[x+dx[j,i],y+dy[j,i]]:=cl;
 end;
 procedure dfs(x,lt:longint);
 var i,j,sum:longint;
 begin
     if x=m+ then
     begin
         for i:= to m do if a[,i]= then exit;
         sum:=;
         for i:= to  do
             for j:= to m do
                 sum:=sum*+a[i,j];
         inc(g[lt][sum]);
         exit;
     end;
     dfs(x+,lt);
     for i:= to  do
         if check(x,i) then
         begin
             fill(x,i,);
             dfs(x+,lt);
             fill(x,i,);
         end;
 end;
 procedure bfs;
 var i,j,l,r,x,y:longint;
 begin
     fillchar(v1,sizeof(v1),false);
     fillchar(v2,sizeof(v2),false);
     l:=; r:=; b[]:=(<<m-)<<m; v1[(<<m-)<<m]:=true;
     while l<=r do
     begin
         x:=b[l];
         for y:= to (<<(m*)-) do
             if (g[x,y]>) and not v1[y] then
             begin
                 v1[y]:=true;
                 inc(r);
                 b[r]:=y;
             end;
         inc(l);
     end;
     l:=; r:=; b[]:=(<<m-)<<m; v2[(<<m-)<<m]:=true;
     while l<=r do
     begin
         x:=b[l];
         for y:= to <<(m*)- do
             if (g[y,x]>) and not v2[y] then
             begin
                 v2[y]:=true;
                 inc(r);
                 b[r]:=y;
             end;
         inc(l);
     end;
     xh:=;
     for i:= to <<(m*)- do
         if v1[i] and v2[i] then
         begin
             inc(xh);
             pos[i]:=xh;
         end;
     for i:= to r do
         for j:= to r do
             mat[pos[b[i]],pos[b[j]]]:=g[b[i],b[j]];
 end;
 operator *(a,b:arr) c:arr; inline;
 var i,j,k:longint;
 begin
     for i:= to xh do
         for j:= to xh do
         begin
             c[i,j]:=;
             for k:= to xh do
                 c[i,j]:=(c[i,j]+a[i,k]*b[k,j]);
             c[i,j]:=c[i,j] mod mo;
         end;
     exit(c);
 end;
 begin
     assign(input,'tile.in');reset(input);
     assign(output,'tile.out');rewrite(output);
     readln(n,m);
     for i:= to <<(*m)- do
     begin
         for j:= to m do
             if i and (<<(m*-j))> then a[,j]:= else a[,j]:=;
         for j:= to m do
             if i and (<<(m-j))> then a[,j]:= else a[,j]:=;
         dfs(,i);
     end;
     {dp[1][(1<<m-1)<<m]:=1;
     for i:=2 to n+1 do
         for j:=0 to 1<<(2*m)-1 do
             for k:=0 to 1<<(2*m)-1 do
                 dp[i][j]:=(dp[i][j]+dp[i-1][k]*g[k][j]) mod mo;}
     bfs;
     for i:= to xh do f[i,i]:=;
     writeln(xh);
     while n> do
     begin
         if n and = then f:=f*mat;
         mat:=mat*mat;
         n:=n>>;
     end;
     writeln(f[pos[(<<m-)<<m]][pos[(<<m-)<<m]]);
     close(input);
     close(output);
 end.

AC 代码2 ：(Orz rpCardinal)

 #include <cstdio>
 #include <cstring>
 #define P 65521

 #ifdef _WIN32
 #define ll "%I64d"
 #else
 #define ll "%lld"
 #endif

 int m,M,ST,N,q[],h[],pos[];
 long long n,g[][],t[][];
 bool b[][],v1[],v2[];

 struct matrix
 {
     long long a[][];
     matrix() {memset(a,,sizeof(a));}
     void one() {for (int i=;i<=N;++i) a[i][i]=;}
     matrix& operator*=(const matrix &B)
     {
         memset(t,,sizeof(t));
         for (int i=;i<=N;++i)
             for (int j=;j<=N;++j)
                 for (int k=;k<=N;++k)
                     t[i][j]=(t[i][j]+a[i][k]*B.a[k][j])%P;
         memcpy(a,t,sizeof(a)); return *this;
     }
 }A,R;

 bool can(int i,int j)
 {return i>=&&i<&&j>=&&j<m&&!b[i][j];}

 bool check(int k,int i)
 {
     switch (k)
     {
         case :return can(,i)&&can(,i+)&&can(,i);
         case :return can(,i)&&can(,i+)&&can(,i);
         case :return can(,i)&&can(,i-)&&can(,i);
         case :return can(,i)&&can(,i-)&&can(,i);
         case :return can(,i)&&can(,i)&&can(,i)&&can(,i+);
         case :return can(,i)&&can(,i)&&can(,i-)&&can(,i+);
         case :return can(,i)&&can(,i)&&can(,i)&&can(,i-);
         case :return can(,i)&&can(,i)&&can(,i-)&&can(,i+);
         default:return ;
     }
 }

 bool fill(int k,int i,int v)
 {
     switch (k)
     {
         case :return b[][i]=b[][i+]=b[][i]=v;
         case :return b[][i]=b[][i+]=b[][i]=v;
         case :return b[][i]=b[][i-]=b[][i]=v;
         case :return b[][i]=b[][i-]=b[][i]=v;
         case :return b[][i]=b[][i]=b[][i]=b[][i+]=v;
         case :return b[][i]=b[][i]=b[][i-]=b[][i+]=v;
         case :return b[][i]=b[][i]=b[][i]=b[][i-]=v;
         case :return b[][i]=b[][i]=b[][i-]=b[][i+]=v;
         default:return ;
     }
 }

 void dfs(int dep,int last)
 {
     if (dep>m)
     {
         for (int i=;i<m;++i) if (!b[][i]) return;
         int next=;
         for (int i=;i;--i)
             for (int j=m-;j>=;--j)
                 next=next*+b[i][j];
         ++g[last][next]; return;
     }
     dfs(dep+,last);
     for (int i=;i<=;++i)
         if (check(i,dep))
         {fill(i,dep,); dfs(dep+,last); fill(i,dep,);}
 }

 void bfs()
 {
     int l=,r=; q[++r]=ST; v1[ST]=;
     while (l<r)
     {
         int x=q[++l];
         for (int y=;y<M;++y)
             if (g[x][y]&&!v1[y]) {v1[y]=; q[++r]=y;}
     }
     l=r=; q[++r]=ST; v2[ST]=;
     while (l<r)
     {
         int x=q[++l];
         for (int y=;y<M;++y)
             if (g[y][x]&&!v2[y]) {v2[y]=; q[++r]=y;}
     }
     for (int i=;i<M;++i)
         if (v1[i]&&v2[i]) {h[++N]=i; pos[i]=N;}
     for (int i=;i<M;++i)
         for (int j=;j<M;++j)
             A.a[pos[i]][pos[j]]=g[i][j];
 }

 void pow(long long b)
 {while (b) {if (b&) R*=A; A*=A; b>>=;}}

 int main()
 {
     freopen("tile.in","r",stdin);
     freopen("tile.out","w",stdout);
     scanf(ll "%d",&n,&m); M=<<(*m); ST=(<<m)-;
     for (int st=;st<M;++st)
     {
         for (int i=;i<m;++i) b[][i]=(st>>i)&;
         for (int i=;i<m;++i) b[][i]=(st>>(i+m))&;
         dfs(,st);
     }
     bfs(); R.one(); pow(n);
     printf(ll "\n",R.a[pos[ST]][pos[ST]]);
     fclose(stdin); fclose(stdout);
     return ;
 }