先预处理出每一行的可行状压值存入mark，并记录该值所含点数sum

处理每相邻两行的状压值是否可行存入 dis

DP方程：dp[i][l][k+sum[l]]+=dp[i-1][j][k];  第i行l状态一共放了k+sum[l]+=上一行j状态放k个的方案数

#include "stdio.h"
#include "string.h"

int b[21];
int n,m,i,j,l,k,cnt;
long long ans,dp[2][101][90];
int dis[101][101],sum[101],mark[101];
int get(int x)
{
    int sum;
    sum=0;
    while (x!=0)
    {
        sum+=x%2;
        x/=2;
    }
    return sum;
}

int judge1(int x)
{
    if (x&(x<<1)) return 0;
    if (x&(x>>1)) return 0;
    return 1;
}

int judge2(int x,int y)
{
    if (x&y) return 0;
    if (x&(y<<1)) return 0;
    if (x&(y>>1)) return 0;
    return 1;
}

int main()
{

    b[0]=1;
    for (i=1;i<=10;i++)
        b[i]=b[i-1]*2;

    while (scanf("%d%d",&n,&m)!=EOF)
    {
        cnt=0;
        for (i=0;i<b[n];i++)
            if (judge1(i)==1)
            {
                cnt++;
                mark[cnt]=i;
                sum[cnt]=get(i);
            }

        memset(dis,0,sizeof(dis));

        for (i=1;i<=cnt;i++)
            for (j=1;j<=cnt;j++)
            if (judge2(mark[i],mark[j])==1)
            dis[i][j]=1;

        memset(dp,0,sizeof(dp));
        dp[0][1][0]=1;

        for (i=1;i<=n;i++)
        {
            memset(dp[i%2],0,sizeof(dp[i%2]));
            for (j=1;j<=cnt;j++)
                for (k=0;k<=m;k++)
                if (dp[1-i%2][j][k])
                    for (l=1;l<=cnt;l++)
                        if (dis[j][l]&&k+sum[l]<=m)
                            dp[i%2][l][k+sum[l]]+=dp[1-i%2][j][k];
        }
        ans=0;
        for (i=1;i<=cnt;i++)
            ans+=dp[n%2][i][m];
        printf("%lld\n",ans);
    }
    return 0;
}

BZOJ 1087 状压DP