SGU 223
题意：给你n*n的矩形，放k个国王，每个国王不能放在别的国王的8连边上，问你有多少种方法
收获：状态DP，因为每行的放置只会影响下一行，然我们就枚举每行的状态和对应的下一行的状态，当两个状态合法时就是可以
转移的时候，然后枚举从第一行到当前行用了多少个，转移一下就行了
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e5+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int n,k; ll dp[16][112][1028]; int dx[] = {0,0,1,-1,-1,-1,1,1}; int dy[] = {1,-1,0,0,-1,1,-1,1}; bool ok(int pre,int now){ int arr[5][25] = {0}; rep(i,1,n+1) arr[1][i] = (pre&(1<<(i-1))); rep(i,1,n+1) arr[2][i] = (now&(1<<(i-1))); rep(i,1,3) { rep(j,1,n+1){ if(arr[i][j]){ int x = i, y = j; rep(t,0,8){ int nx = x + dx[t],ny = y + dy[t]; if(arr[nx][ny]) return false; } } } } return true; } int bit(int s){ bitset<10> ans(s); return ans.count(); } int main(){ scanf("%d%d",&n,&k); dp[0][0][0] = 1; rep(i,0,n) rep(pre,0,1<<n) rep(now,0,1<<n) if(ok(pre,now)){ rep(t,0,k-bit(now)+1) dp[i+1][bit(now)+t][now] += dp[i][t][pre]; } ll ans = 0; rep(i,0,1<<n) ans += dp[n][k][i]; printf("%lld\n",ans); return 0; }