Submission #1336028

#TimeUsernameProblemLanguageResultExecution timeMemory
1336028YSH2020Catfish Farm (IOI22_fish)C++20
3 / 100
49 ms7260 KiB
#include "fish.h"

#include <bits/stdc++.h>
using namespace std;
long long max_weights(int N, int M, vector<int> X, vector<int> Y, vector<int> W) {
	long long ans = 0; for (int i =0; i < M; i++) ans += W[i];
	return ans;
}
//long long max_weights(int N, int M, std::vector<int> X, std::vector<int> Y,
                      //std::vector<int> W) {
	//int weights[N][N+1]; memset(weights,0,sizeof(weights));
	//for (int i = 0; i < M; i++) {
		//weights[X[i]][Y[i]+1]=W[i];
	//}
	//int pref[N][N+1];
	//for (int i = 0; i < N; i++) {
		//pref[i][0]=0;
		//for (int j = 1; j <= N; j++) pref[i][j]=pref[i][j-1]+weights[i][j];
	//}
	//int dp[N][N+1];
	////the thing is, we first do dp on N=1. this is to be more convenient i guess?
	//for (int i = 0; i <= N; i++) {
		////if this is higher than the previous, then of course optimal to take everything above,
		//dp[0][i][0]=pref[1][i];
	//}
	//int store1[N][N+1], store2[N][N+1], store3[N][N+1], store4[N][N+1];
	//for (int i = 1; i < N; i++) {
		//for (int j = 0; j <= N; j++) {
			////left height
			////impact: j<=x -> dp[i+1][x][0]=max(dp[i+1][x][0], dp[i][j][0]
			////case 1: j > k, j > l
			//dp[i][j] = store1[i-1][j]+pref[i+1][j]+pref[i-1][j];
			////case 2: j < k, j < l
			//dp[i][j]=max(dp[i][j],store2[i-1][j]+pref[i+1][j]);
			////case 3: j < k, j > l
			//dp[i][j]=max(dp[i][j],store3[i-1][j]+pref[i-1][j]-pref[i][j]+pref[i+1][j]);
			////case 4: j < k, j < l
			//dp[i][j]=max(dp[i][j],store4[i-1][j]-pref[i][j]+pref[i+1][j]);
			////why ?????????????? IM SO SAD RNNNN
		//}
		////now we update the stores
		//store1[i][j] = max(dp[i-1][k][l]-pref[i-1][k]) //where j > k, j > l
	//}
//}

//???
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