bzoj1087

题解：

状压dp

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=;
int n,m,cnt[N],c1[N],c2[N][N];
ll ans,f[][N][N];
void pre()
{
    int s;
    for (int i=;i<<<n;i++)
     if ((i&(i>>))==)
      {
        s=;
        for (int x=i;x;x>>=)s+=(x&);
        cnt[i]=s;c1[i]=;
      }
    for (int i=;i<<<n;i++)
     if (c1[i])
      for (int j=;j<<<n;j++)
       if (c1[j])
        if (((i&j)==)&&((i&(j>>))==)&&((j&(i>>))==))c2[i][j]=;
}
int main()
{
    scanf("%d%d",&n,&m);
    pre();
    for (int i=;i<<<n;i++)
     if (c1[i])f[][cnt[i]][i]=;
    for (int j=;j<n;j++)
     for (int k=;k<<<n;k++)
      if (c1[k])
       for (int i=;i<<<n;i++)
        if (c1[i])
         if (c2[k][i])
          for (int p=cnt[k];p+cnt[i]<=m;p++)f[j+][p+cnt[i]][i]+=f[j][p][k];
    ll ans=;
    for (int i=;i<<<n;i++)ans+=f[n][m][i];
    printf("%lld",ans);
    return ;
}