// WIE
#ifndef __SIZEOF_INT128__
#define __SIZEOF_INT128__
#endif
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace std;
using namespace chrono;
using namespace __gnu_pbds;
template <typename T> using oset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define rep(i, p, k) for(int i(p); i < k; ++i)
#define per(i, p, k) for(int i(p); i > k; --i)
#define sz(x) (int)x.size()
#define sc static_cast
typedef long long ll;
typedef long double ld;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef __int128_t lll;
//#define int ll
template <typename T = int> using par = std::pair <T, T>;
#define fir first
#define sec second
#define test int _number_of_tests(in()); while(_number_of_tests--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define pb emplace_back
struct Timer {
string name{""};
time_point<high_resolution_clock> end, start{high_resolution_clock::now()};
duration<float, std::milli> dur;
Timer() = default;
Timer(string name): name(name) {}
~Timer() {
end = high_resolution_clock::now(); dur= end - start;
cout << "@" << name << "> " << dur.count() << " ms" << '\n';
}
};
template <typename T = int> inline T in()
{
T x;
std::cin >> x;
return x;
}
std::string yn(bool b)
{
if(b) return "YES\n";
else return "NO\n";
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par);
template <typename T> std::ostream& operator<< (std::ostream& out, const std::vector <T>& wek)
{
for(const auto& i : wek)out << i << ' ';
return out;
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par)
{
out << '{'<<par.first<<", "<<par.second<<"}";
return out;
}
#define show(x) cerr << #x << " = " << x << '\n';
ll dp[9][(1<<8) + 3][9][2];
// dp[a][b][c][d] - suma liczb mozliwych ruchow dla c pionkow w pierwszych a rzedach z ustawieniem b w ostatnim i parzystosci d.
ll moz[9][(1<<8) + 3][9][2];
// moz[a][c][d] - na ile sposobow mozna po pierwszych a liniach rozstawic c pionkow z parzystoscia d.
bool tab[9][9];
bool ta2[9][9];
std::int32_t main()
{
std::cin.tie(nullptr); std::cout.tie(nullptr); std::ios_base::sync_with_stdio(0);
cout << setprecision(15) << fixed;
int n(in()), m(in());
moz[0][0][0][0] = 1;
rep(a, 1, n+1)rep(b, 0, 1<<m)rep(c, 0, 9)rep(d, 0, 2){
int pc(c);
bool pd(d);
rep(i, 0, m)if(b&1<<i){
--pc;
pd ^= (i + a) % 2;
}
if(pc < 0)continue;
rep(pb, 0, 1<<m){
moz[a][b][c][d] += moz[a-1][pb][pc][pd];
dp[a][b][c][d] += dp[a-1][pb][pc][pd];
ll s(0);
rep(i, 0, m-1)s += !!(b&1<<i) ^ !!(b&1<<i+1);
if(a > 1)rep(i, 0, m) s += !!(b&1<<i) ^ !!(pb&1<<i);
dp[a][b][c][d] += s * moz[a-1][pb][pc][pd];
}
}
bool d(0);
int c(0);
rep(i, 1, n+1)rep(j, 1, m+1){
tab[i][j] = in<char>() == 'O';
if((i + j) % 2 == 0)d ^= tab[i][j];
c += tab[i][j];
}
rep(i, 1, n+1)rep(j, 1, m+1){
ta2[i][j] = in<char>() == 'O';
if((i + j) % 2 == 0)d ^= ta2[i][j];
}
if(d){
cout << "0\n";
return 0;
}
ll s(0), t(0);
rep(b, 0, 1<<m)s += dp[n][b][c][d];
rep(i, 1, n+1)rep(j, 1, m+1)if(ta2[i][j]){
t += i > 1 && !ta2[i-1][j];
t += i < n && !ta2[i+1][j];
t += j > 1 && !ta2[i][j-1];
t += j < m && !ta2[i][j+1];
}
cout << (long double)(t)/s << '\n';
return 0;
}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 | // WIE #ifndef __SIZEOF_INT128__ #define __SIZEOF_INT128__ #endif #pragma GCC optimize("Ofast") #include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/priority_queue.hpp> using namespace std; using namespace chrono; using namespace __gnu_pbds; template <typename T> using oset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>; #define rep(i, p, k) for(int i(p); i < k; ++i) #define per(i, p, k) for(int i(p); i > k; --i) #define sz(x) (int)x.size() #define sc static_cast typedef long long ll; typedef long double ld; typedef unsigned int uint; typedef unsigned long long ull; typedef __int128_t lll; //#define int ll template <typename T = int> using par = std::pair <T, T>; #define fir first #define sec second #define test int _number_of_tests(in()); while(_number_of_tests--) #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define pb emplace_back struct Timer { string name{""}; time_point<high_resolution_clock> end, start{high_resolution_clock::now()}; duration<float, std::milli> dur; Timer() = default; Timer(string name): name(name) {} ~Timer() { end = high_resolution_clock::now(); dur= end - start; cout << "@" << name << "> " << dur.count() << " ms" << '\n'; } }; template <typename T = int> inline T in() { T x; std::cin >> x; return x; } std::string yn(bool b) { if(b) return "YES\n"; else return "NO\n"; } template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par); template <typename T> std::ostream& operator<< (std::ostream& out, const std::vector <T>& wek) { for(const auto& i : wek)out << i << ' '; return out; } template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par) { out << '{'<<par.first<<", "<<par.second<<"}"; return out; } #define show(x) cerr << #x << " = " << x << '\n'; ll dp[9][(1<<8) + 3][9][2]; // dp[a][b][c][d] - suma liczb mozliwych ruchow dla c pionkow w pierwszych a rzedach z ustawieniem b w ostatnim i parzystosci d. ll moz[9][(1<<8) + 3][9][2]; // moz[a][c][d] - na ile sposobow mozna po pierwszych a liniach rozstawic c pionkow z parzystoscia d. bool tab[9][9]; bool ta2[9][9]; std::int32_t main() { std::cin.tie(nullptr); std::cout.tie(nullptr); std::ios_base::sync_with_stdio(0); cout << setprecision(15) << fixed; int n(in()), m(in()); moz[0][0][0][0] = 1; rep(a, 1, n+1)rep(b, 0, 1<<m)rep(c, 0, 9)rep(d, 0, 2){ int pc(c); bool pd(d); rep(i, 0, m)if(b&1<<i){ --pc; pd ^= (i + a) % 2; } if(pc < 0)continue; rep(pb, 0, 1<<m){ moz[a][b][c][d] += moz[a-1][pb][pc][pd]; dp[a][b][c][d] += dp[a-1][pb][pc][pd]; ll s(0); rep(i, 0, m-1)s += !!(b&1<<i) ^ !!(b&1<<i+1); if(a > 1)rep(i, 0, m) s += !!(b&1<<i) ^ !!(pb&1<<i); dp[a][b][c][d] += s * moz[a-1][pb][pc][pd]; } } bool d(0); int c(0); rep(i, 1, n+1)rep(j, 1, m+1){ tab[i][j] = in<char>() == 'O'; if((i + j) % 2 == 0)d ^= tab[i][j]; c += tab[i][j]; } rep(i, 1, n+1)rep(j, 1, m+1){ ta2[i][j] = in<char>() == 'O'; if((i + j) % 2 == 0)d ^= ta2[i][j]; } if(d){ cout << "0\n"; return 0; } ll s(0), t(0); rep(b, 0, 1<<m)s += dp[n][b][c][d]; rep(i, 1, n+1)rep(j, 1, m+1)if(ta2[i][j]){ t += i > 1 && !ta2[i-1][j]; t += i < n && !ta2[i+1][j]; t += j > 1 && !ta2[i][j-1]; t += j < m && !ta2[i][j+1]; } cout << (long double)(t)/s << '\n'; return 0; } |
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