Submission #1172465

#	Time	Username	Problem	Language	Result	Execution time	Memory
1172465		thelegendary08	메기 농장 (IOI22_fish)	C++17	0 / 100	948 ms	2162688 KiB

fish

#include "fish.h"
// #include "grader.cpp"
#include<bits/stdc++.h>
#define int long long
#define vi vector<int>
#define pb push_back
#define f0r(i,n) for(int i = 0; i<n; i++)
#define FOR(i, k, n) for(int i = k; i<n; i++)
#define vout(v) for(auto u : v)cout<<u<<' '; cout<<'\n';

using namespace std;
vector<vector<int>>s;
int sum(int col, int l, int r){
	if(l > r)return 0;
	if(l == 0)return s[col][r];
	else return s[col][r] - s[col][l-1];
}
int max_weights(signed n, signed m, std::vector<signed> x, std::vector<signed> y,
                      std::vector<signed> w) {
	vector<vi>grid(n, vi(n));
	f0r(i,m){
		grid[x[i]][y[i]] = w[i];
	}
	vector<vi>s(n, vi(n));
	f0r(i, n){
		s[i][0] = grid[i][0];
		FOR(j, 1, n)s[i][j] = s[i][j-1] + grid[i][j];
	}
	::s = s;
	vector<vector<vi>>dp(n, vector<vi>(n+1, vi(2))); // index, prev, 0 = inc, 1 = dec, 0 goes in inc
	// if currently increasing, prev must be increasing. if currently decreasing, prev can be anything.
	vi zerotake(n); //for this dp[i][0][0], how many were already taken in the current column?
	FOR(i, 1, n){
		f0r(j,n+1){
			f0r(k, 2){
				if(k == 0){
					if(j == 0){
						//current one is empty
						f0r(l, n+1){
							f0r(o, 2)dp[i][j][k] = max(dp[i][j][k], dp[i-1][l][o] + sum(i, 0, l-1));
						}
						zerotake[i] = dp[i][j][k];
					}
					else{
						//current one is increasing
						f0r(l, j){
							if(l != 0)dp[i][j][k] = max(dp[i][j][k], dp[i-1][l][0] + sum(i-1, l, j-1));
							else dp[i][j][k] = max(dp[i][j][k], dp[i-1][l][0] + max(0LL, sum(i-1, l, j-1) - zerotake[i-1]));
							
						}
					}
				}
				else{
					//current one is decreasing, previous length must be higher than current
					FOR(l, j+1, n+1){
						f0r(o,2)dp[i][j][k] = max(dp[i][j][k], dp[i-1][l][o] + sum(i, j, l-1));
					}
				}
			}
		}
	}
	/*
	f0r(i,n+1){
		f0r(j, 2){
			cout<<dp[1][i][j]<<' ';
		}
		cout<<'\n';
	}
	*/
	int ans = 0;
	f0r(i,n+1)f0r(j,2)ans = max(ans, dp[n-1][i][j]);
	return ans;
}

• Subtask 1: 0 / 3
• Subtask 2: 0 / 6
• Subtask 3: 0 / 9
• Subtask 4: 0 / 14
• Subtask 5: 0 / 21
• Subtask 6: 0 / 17
• Subtask 7: 0 / 14
• Subtask 8: 0 / 16