Submission #1256580

#	Time	Username	Problem	Language	Result	Execution time	Memory
1256580		lioow	Maja (COCI18_maja)	C++20	0 / 110	1 ms	584 KiB

maja

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define pb push_back
#define int long long
#define repp(i,x,n) for(int i=x;i<=n;i++)
#define rep(i,x,n) for(int i=x;i>=n;i--)
#define r0 return 0
#define fi first
#define se second
#define liow ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define jelek cout<<"jelek"<<endl
#define pii pair<int,int>
#define all(v) v.begin(),v.end()
#define tp tuple<int,int,int>
#define fl fflush(stdout)
#define ld long double
#define p5 pair<int,pair<pair<int,int>,pair<int,int>>>
#pragma GCC optimize ("O2")
#pragma GCC optimize ("unroll-loops")  
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<pair<int,int>, null_type,  less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
//const int mod=1e9+7;
const int SQMAX=635,INF=1e18;
const int mod=998244353;
//const int MOD=1e6+3;
mt19937_64 rng((unsigned int) chrono::steady_clock::now().time_since_epoch().count());
pii dr[8]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int dxx[4]={1,1,-1,-1};
int dyy[4]={1,-1,-1,1};
int mul(int x,int y){return (x%mod*y%mod)%mod;}
int expo(int b,int e){
	if(e==0) return 1;
	int tmp=expo(b,e/2);
	if(e%2) return mul(tmp,mul(tmp,b));
	else return mul(tmp,tmp);
}
const int maxn=2e5;

void solve(){
	int n,m,a,b,K;cin>>n>>m>>a>>b>>K;
	int aa[n+2][m+2];
	repp(i,1,n){
		repp(j,1,m){
			cin>>aa[i][j];
		}
	}
	int L=5000;
	int dp[n+2][m+2][2];
	repp(i,0,n+1){
		repp(j,0,m+1){
			repp(k,0,1) dp[i][j][k]=-INF;
		}
	}
	dp[a][b][0]=0;
	int ans=0;
	repp(k,1,L){
		if(k*2>K) break;
		int id=k%2;
		repp(i,1,n){
			repp(j,1,m){
				int maks=0;
				repp(ii,0,3){
					if(!(1<=i+dx[ii] && i+dx[ii]<=n && 1<=j+dy[ii] && j+dy[ii]<=n)) continue;
					maks=max(maks,aa[i+dx[ii]][j+dy[ii]]);
					dp[i][j][id]=max(dp[i][j][id],dp[i+dx[ii]][j+dy[ii]][id^1]+aa[i][j]);
				}
				ans=max(ans,2*dp[i][j][id]+(K-2*k)/2*(maks+aa[i][j])-aa[i][j]);
			}
		}
		repp(i,1,n){
			repp(j,1,m) dp[i][j][id^1]=-INF;
		}
	}
	cout<<ans<<endl;
}

signed main(){
  liow;
  int t=1;
  cin>>t;
  while(t--){
  	solve();
  }
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 11
• Subtask 5: 0 / 11
• Subtask 6: 0 / 11
• Subtask 7: 0 / 11
• Subtask 8: 0 / 11
• Subtask 9: 0 / 11
• Subtask 10: 0 / 11