今日SGU 5.23

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
",ans);
    return 0;
}