题解 v

传送门

考场上只会爆搜……觉得重复状态其实有很多但不知道怎么记忆化，结果……

• 对于类似这样n不算太小但只有二三十，而重复状态极多的题其实也是可以跑状压/记搜的，状态可以开map存

然后就是爆搜，就没什么了……

这题\(k=n-1\)其实有特解……白优化了半天hash 考虑\(k=n-1\)其实就是有一个点可以任选

upd: 其实不用特解，移位卡常可以过

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100
#define ll long long
#define ld long double
#define usd unsigned
#define ull unsigned long long
//#define int long long
#define max(a, b) ((a)>(b)?(a):(b))
#define reg register int

inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, k;
char s[N];

namespace force{
	int v[N];
	bool vis[N];
	double dfs(int u, int sum) {
		//cout<<"dfs "<<u<<' '<<sum<<endl;
		if (u>k) return 1.0*sum;
		int lim=n-u+1, bkpa, bkpb, cnt;
		int a, b;
		double ans=0; double t1, t2;
		for (int i=1; i<=lim; ++i) {
			cnt=0; for (int j=1; j<=n; ++j) {if (!vis[j]) {if (++cnt==i) {vis[j]=1; a=v[j]; bkpa=j;}}}
			t1=dfs(u+1, sum+(a==0));
			//cout<<a<<' '<<bkpa<<endl;
			vis[bkpa]=0;
			cnt=0; for (int j=n; j; --j) {if (!vis[j]) {if (++cnt==i) {vis[j]=1; b=v[j]; bkpb=j;}}}
			t2=dfs(u+1, sum+(b==0));
			vis[bkpb]=0;
			//cout<<"u, i: "<<u<<' '<<i<<' '<<max(t1, t2)<<endl;
			ans+=max(t1, t2)/(1.0*lim);
		}
		return ans;
	}
	void solve() {
		for (int i=1; i<=n; ++i)
			if (s[i]=='W') v[i]=0;
			else v[i]=1;
		//cout<<1<<endl;
		printf("%.10lf\n", dfs(1, 0));
		exit(0);
	}
}

namespace task2{
	void solve() {
		int cnt=0;
		for (int i=1; i<=n; ++i) if (s[i]=='W') ++cnt;
		printf("%.10lf\n", double(cnt));
		exit(0);
	}
}

namespace task3{
	void solve() {
		int cntw=0;
		for (int i=1; i<=n; ++i) if (s[i]=='W') ++cntw;
		if (!cntw) {puts("0.0000000000"); exit(0);}
		if (cntw==n) {printf("%.10lf\n", double(k)); exit(0);}
		if (cntw==n-1 && n%2==0) {printf("%.10lf\n", double(k)); exit(0);}
		if (cntw==n-1 && n%2) {printf("%.10lf\n", double(k-1)/double(k)); exit(0);}
		if (k==n-1) {printf("%.10lf\n", double(min(cntw, n-1))); exit(0);}
		force::solve();
	}
}

namespace task{
	//unordered_map<pair<int, int>, double> mp;
	struct ele{int a, b; ele(){} ele(int a_, int b_):a(a_),b(b_){}};
	inline bool operator == (ele a, ele b) {return a.a==b.a&&a.b==b.b;}
	struct hush_table{
		static const int SIZE=50000100;
		int head[SIZE], size;
		//int cnt[SIZE];
		struct edge{double dat; ele p; int next;}e[5000010];
		inline bool find(ele q) {
			ll t=(998244353ll*q.a*q.b+(q.a))%SIZE;
			//cout<<"t: "<<t<<endl;
			for (int i=head[t]; i; i=e[i].next)
				if (q==e[i].p) return 1;
			return 0;
		}
		inline double operator [] (ele q) {
			ll t=(998244353ll*q.a*q.b+(q.a))%SIZE;
			//cout<<"t: "<<t<<endl;
			for (int i=head[t]; i; i=e[i].next)
				if (q==e[i].p) return e[i].dat;
		}
		inline void add(ele q, double dat) {
			//cout<<"size: "<<size<<endl;
			ll t=(998244353ll*q.a*q.b+(q.a))%SIZE;
			//++cnt[t];
			//cout<<"t: "<<t<<endl;
			edge* k=&e[++size]; k->dat=dat; k->p=q; k->next=head[t]; head[t]=size;
			//if (size > 10000010) cout<<"error"<<endl;
		}
		void check() {
			//int maxn=0;
			//for (int i=0; i<SIZE; ++i) maxn=max(maxn, cnt[i]);
			//cout<<"maxn: "<<maxn<<endl;
			//cout<<"size: "<<size<<endl;
		}
	}mp;
	int v[N];
	double dfs(int u, int s) {
		//cout<<"dfs "<<u<<' '<<bitset<10>(s)<<endl;
		if (u>k) return 0.0;
		ele p(u, s);
		//cout<<"go to find"<<endl;
		if (mp.find(p)) return mp[p];
		//cout<<"not return "<<endl;
		int lim=n-u+1, cnt;
		int a, b, s2, s3;
		double ans=0; double t1, t2;
		for (reg i=1; i<=lim/2; ++i) {
			a=(s&(1<<(i-1)))?1:0;
			cnt=s2=0;
			#if 0
			for (reg j=0; j<lim; ++j)
				if (j!=i-1) {
					if (s&(1<<j)) s2|=(1<<cnt);
					++cnt;
				}
			#endif
			s2=((s>>i)<<(i-1));
			s2|=(s&((1<<(i-1))-1));
			//cout<<"s2, s3: "<<bitset<5>(s2)<<' '<<bitset<5>(s3)<<endl;
			//assert(s2==s3);
			t1=dfs(u+1, s2)+(a==0);

			b=(s&(1<<(lim-i)))?1:0;
			cnt=s2=0;
			#if 0
			for (reg j=0; j<lim; ++j)
				if (j!=lim-i) {
					if (s&(1<<j)) s2|=(1<<cnt); //, cout<<"1<<"<<(cnt)<<endl;
					++cnt;
				}
			#endif
			s2=((s>>(lim-i+1))<<(lim-i));
			s2|=(s&((1<<(lim-i))-1));
			//cout<<"s2, s3: "<<bitset<5>(s2)<<' '<<bitset<5>(s3)<<endl;
			//assert(s2==s3);
			t2=dfs(u+1, s2)+(b==0);

			ans+=max(t1, t2)/(0.5*lim);
		}
		if (lim&1) {
			int i=lim/2+1;
			a=(s&(1<<(i-1)))?1:0;
			cnt=s2=0;
			#if 0
			for (reg j=0; j<lim; ++j)
				if (j!=i-1) {
					if (s&(1<<j)) s2|=(1<<cnt);
					++cnt;
				}
			#endif
			s2=((s>>i)<<(i-1));
			s2|=(s&((1<<(i-1))-1));
			t1=dfs(u+1, s2)+(a==0);
			ans+=t1/(1.0*lim);
		}
		//cout<<"go to add"<<endl;
		mp.add(p, ans);
		//cout<<"return "<<endl;
		return ans;
	}
	void solve() {
		int ss=0;
		for (int i=1; i<=n; ++i)
			if (s[i]=='W') v[i-1]=0;
			else v[i-1]=1, ss|=(1<<(i-1));
		//cout<<1<<endl;
		printf("%.10lf\n", dfs(1, ss));
		mp.check();
		exit(0);
	}
}

signed main()
{
	#ifdef DEBUG
	freopen("1.in", "r", stdin);
	#endif

	n=read(); k=read();
	scanf("%s", s+1);
	if (k==0) {puts("0.0000000000"); return 0;}
	if (k==n || (0&&k==n-1)) task2::solve();
	//if (n>=10) task3::solve();
	//force::solve();
	task::solve();

	return 0;
}