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=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 1000010
#define ul unsigned int
#define S 64
int n,m,p,k;
bool flag[][];
struct matrix
{
    int n;ul a[S][S];
    matrix operator *(const matrix&b) const
    {
        matrix c;c.n=n;memset(c.a,,sizeof(c.a));
        for (int i=;i<n;i++)
            for (int j=;j<S;j++)
                for (int k=;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=;i<;i++)
        for (int j=;j<p;j++)
        flag[i][j]=read();
    flag[][k]=;
    f.n=;f.a[][]=;
    a.n=<<m;
    for (int i=;i<(<<m);i++)
        for (int j=;j<(<<m);j++)
        {
            a.a[i][j]=;
            for (int v=;v<m;v++)
            if (i&(<<v))
                for (int x=max(,v-k);x<min(m,v+p-k);x++)
                {
                    if (flag[][x-(v-k)]&&(i&(<<x))) a.a[i][j]=;
                    if (flag[][x-(v-k)]&&(j&(<<x))) a.a[i][j]=;
                }
            for (int v=;v<m;v++)
            if (j&(<<v))
                for (int x=max(,v-k);x<min(m,v+p-k);x++)
                {
                    if (flag[][x-(v-k)]&&(j&(<<x))) a.a[i][j]=;
                    if (flag[][x-(v-k)]&&(i&(<<x))) a.a[i][j]=;
                }
        }
    for (;n;n>>=,a=a*a) if (n&) f=f*a;
    ul ans=;
    for (int i=;i<(<<m);i++) ans+=f.a[][i];
    cout<<ans;
    return ;
}