[BZOJ4000][TJOI2015]棋盘(状压DP+矩阵快速幂)

题意极其有毒，注意给的行列都是从0开始的。

状压DP，f[i][S]表示第i行状态为S的方案数，枚举上一行的状态转移。$O(n2^{2m})$

使用矩阵加速，先构造矩阵a[S1][S2]表示上一行为S1是下一行是否能为S2，快速幂加速后得解。$O(2^{3m}m^2+2^{3m}\log n)$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef unsigned int ul;
using namespace std;

const int N=70;
ul ans;
int n,m,p,k,ed,x,tot;
struct P{ int x,y; }d[110];

struct Mat{
ul a

;
ul* operator [](int x){ return a[x]; }
Mat (){ memset(a,0,sizeof(a)); }
}a,res;

Mat mul(Mat a,Mat b){
Mat c;
rep(i,0,ed) rep(j,0,ed) rep(k,0,ed) c[i][k]+=a[i][j]*b[j][k];
return c;
}

Mat ksm(Mat a,int b){
Mat res;
rep(i,0,ed) res[i][i]=1;
for (; b; a=mul(a,a),b>>=1)
if (b & 1) res=mul(res,a);
return res;
}

int main(){
freopen("bzoj4000.in","r",stdin);
freopen("bzoj4000.out","w",stdout);
scanf("%d%d%d%d",&n,&m,&p,&k); k++; ed=(1<<m)-1;
rep(i,1,3) rep(j,1,p){
scanf("%d",&x);
if (x) d[++tot]=(P){i-2,j-k};
}
rep(S1,0,ed) rep(S2,0,ed){
bool flag=0;
rep(i,0,m-1) if (S1&(1<<i))
rep(j,0,m-1) if (S2&(1<<j))
rep(k,1,tot) if ((d[k].x==1 && d[k].y==j-i) || (d[k].x==-1 && d[k].y==i-j))
{ flag=1; break; }
rep(i,0,m-2) if (S2&(1<<i))
rep(j,i+1,m-1) if (S2&(1<<j))
rep(k,1,tot) if (d[k].x==0 && (d[k].y==j-i || d[k].y==i-j))
{ flag=1; break; }
a[S1][S2]=!flag;
}
res[0][0]=1; res=mul(res,ksm(a,n));
rep(i,0,ed) ans+=res[0][i];
printf("%u\n",ans);
return 0;
}