AC通道：http://www.lydsy.com/JudgeOnline/problem.php?id=1087
【题解】
用f[i][j][k]表示前i行放了j个棋子且第i行的状态为k的方案数。
vis[i]表示状态i是否合法，check[i][j]表示状态i，j是否可以相邻。
详见代码：
/*************
bzoj 1087
by chty
2016.11.15
*************/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
#define FILE "read"
#define up(i,j,n) for(ll i=j;i<=n;i++)
namespace INIT{
char buf[1<<15],*fs,*ft;
inline char getc() {return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;}
inline ll read() {
ll x=0,f=1; char ch=getc();
while(!isdigit(ch)) {if(ch=='-') f=-1; ch=getc();}
while(isdigit(ch)) {x=x*10+ch-'0'; ch=getc();}
return x*f;
}
}using namespace INIT;
ll n,m,ans,vis[1<<9],sum[1<<9],check[1<<9][1<<9],f[11][121][1<<9];
void pre(){
up(i,0,(1<<n)-1)if(!(i&(i<<1))){vis[i]=1; for(ll x=i;x;x>>=1) sum[i]+=(x&1);}
up(i,0,(1<<n)-1){
if(!vis[i]) continue;
up(j,0,(1<<n)-1){
if(!vis[j]||(i&j)||(i&(j>>1))||(j&(i>>1))) continue;
check[i][j]=1;
}
}
}
int main(){
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
n=read(); m=read();
pre(); up(i,0,(1<<n)-1) if(vis[i]) f[1][sum[i]][i]=1;
up(i,2,n)up(j,0,(1<<n)-1){
if(!vis[j]) continue;
up(k,0,(1<<n)-1){
if(!vis[k]||!check[j][k]) continue;
up(p,sum[k],m-sum[j]) f[i][p+sum[j]][j]+=f[i-1][p][k];
}
}
up(i,0,(1<<n)-1) ans+=f[n][m][i];
printf("%lld
",ans);
return 0;
}