BZOJ4000 TJOI2015棋盘（状压dp+矩阵快速幂）

　　显然每一行棋子的某种放法是否合法只与上一行有关，状压起来即可。然后n稍微有点大，矩阵快速幂即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
int x=0,f=1;char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
#define N 1000010
#define ul unsigned int
#define S 64
int n,m,p,k;
bool flag[3][6];
struct matrix
{
int n;ul a[S][S];
matrix operator *(const matrix&b) const
{
matrix c;c.n=n;memset(c.a,0,sizeof(c.a));
for (int i=0;i<n;i++)
for (int j=0;j<S;j++)
for (int k=0;k<S;k++)
c.a[i][j]+=a[i][k]*b.a[k][j];
return c;
}
}f,a;
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj4000.in","r",stdin);
freopen("bzoj4000.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
n=read(),m=read(),p=read(),k=read();
for (int i=0;i<3;i++)
for (int j=0;j<p;j++)
flag[i][j]=read();
flag[1][k]=0;
f.n=1;f.a[0][0]=1;
a.n=1<<m;
for (int i=0;i<(1<<m);i++)
for (int j=0;j<(1<<m);j++)
{
a.a[i][j]=1;
for (int v=0;v<m;v++)
if (i&(1<<v))
for (int x=max(0,v-k);x<min(m,v+p-k);x++)
{
if (flag[1][x-(v-k)]&&(i&(1<<x))) a.a[i][j]=0;
if (flag[2][x-(v-k)]&&(j&(1<<x))) a.a[i][j]=0;
}
for (int v=0;v<m;v++)
if (j&(1<<v))
for (int x=max(0,v-k);x<min(m,v+p-k);x++)
{
if (flag[1][x-(v-k)]&&(j&(1<<x))) a.a[i][j]=0;
if (flag[0][x-(v-k)]&&(i&(1<<x))) a.a[i][j]=0;
}
}
for (;n;n>>=1,a=a*a) if (n&1) f=f*a;
ul ans=0;
for (int i=0;i<(1<<m);i++) ans+=f.a[0][i];
cout<<ans;
return 0;
}