标签：int 题解 long ele double SIZE define
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;
}
标签：int,题解,long,ele,double,SIZE,define 来源： https://www.cnblogs.com/narration/p/15029829.html
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