結果
問題 | No.2094 Symmetry |
ユーザー |
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提出日時 | 2022-10-07 22:14:28 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 159 ms / 2,000 ms |
コード長 | 2,123 bytes |
コンパイル時間 | 4,999 ms |
コンパイル使用メモリ | 258,332 KB |
最終ジャッジ日時 | 2025-02-07 23:29:49 |
ジャッジサーバーID (参考情報) |
judge4 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 34 |
ソースコード
#include <bits/stdc++.h>using namespace std;#include <atcoder/all>using namespace atcoder;#define rep(i, n) for (int i = 0; i < (long long)(n); i++)#define rep2(i, m ,n) for (int i = (m); i < (long long)(n); i++)#define REP(i, n) for (int i = 1; i < (long long)(n); i++)typedef long long ll;#define updiv(N,X) (N + X - 1) / X#define l(n) n.begin(),n.end()#define YesNo(Q) Q==1?cout<<"Yes":cout<<"No"using P = pair<int, int>;using mint = modint998244353;const int MOD = 998244353LL;const ll INF = 999999999999LL;vector<long long> fact, fact_inv, inv;/* init_nCk :二項係数のための前処理計算量:O(n)*/void init_nCk(int SIZE) {fact.resize(SIZE + 5);fact_inv.resize(SIZE + 5);inv.resize(SIZE + 5);fact[0] = fact[1] = 1;fact_inv[0] = fact_inv[1] = 1;inv[1] = 1;for (int i = 2; i < SIZE + 5; i++) {fact[i] = fact[i - 1] * i % MOD;inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD;}}/* nCk :MODでの二項係数を求める(前処理 int_nCk が必要)計算量:O(1)*/long long nCk(int n, int k) {assert(!(n < k));assert(!(n < 0 || k < 0));return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD;}long long modpow(long long a, long long n, long long mod) {long long res = 1;while (n > 0) {if (n & 1) res = res * a % mod;a = a * a % mod;n >>= 1;}return res;}int main() {int n;ll b;cin>>n>>b;vector<string>s(2*n);rep(i,2*n){cin>>s[i];}ll v[2*n][2*n];rep(i,2*n){rep(j,2*n){cin>>v[i][j];}}ll ls = 0;ll rs=0;rep(i,2*n){rep(j,2*n){if(s[i][j]=='#'){if(n>j){ls++;}else{rs++;}}}}ll ans=0;vector<ll>aa(4*n*n);rep(i,2*n){rep(j,2*n){aa[i*2*n+j]=v[i][j];}}sort(l(aa));reverse(l(aa));rep(i,ls+rs){ans += aa[i];}if((ls+rs)%2==0){ll anss = 0;vector<ll>aaa(2*n*n);rep(i,2*n){rep(j,n){aaa[i*n+j]=v[i][j]+v[i][n*2-j-1];}}sort(l(aaa));reverse(l(aaa));rep(i,(ls+rs)/2){anss += aaa[i];}cout << max(ans,anss+b)<<endl;}else{cout<<ans<<endl;}}