제출 #1180615

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1180615		madamadam3	메기 농장 (IOI22_fish)	C++20	3 / 100	53 ms	14008 KiB

fish

#include "fish.h"
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
using vi = vector<int>;
using vl = vector<ll>;
#define sz(x) (int) (x).size()
#define all(x) (x).begin(), (x).end()
#define pb push_back

/*
  Time log:
  - first start: 1048am
  - 3/100: 1100am
  - first stop: 1128am
  - second start: 337pm
  - 416pm finally stopped being stupid and got 9/100 (+6 points)
  - second end:
  - overall time used: 
*/

/*
  NxN grid of cells
  grid[c][r] = the cell at column c row r
  grid[0][r] is the westernmost point, and grid[N-1][r] is the easternmost point
  grid[c][0] = southernmost point, grid[c][N-1] northermost point

  M catfishes
  X[i], Y[i] for i < M are column and row of fish i
  W[i] = weight of fish i
  a catfish can be caught if grid[ci][ri] is empty and grid[ci-1][ri] or grid[ci+1][ri] has a pier tile
  a pier extends from grid[cP][0] to grid[cP][rP]
*/

int N, M;
vi X, Y; vector<ll> W;

bool cmp(int a, int b) { // compare two fish in the same col
  return Y[a] < Y[b];
}

ll max_weights(int _N, int _M, vi _X, vi _Y, vi _W) {
  N = _N; M = _M; X = _X; Y = _Y; for (int i = 0; i < M; i++) W.pb(ll(_W[i]));

  // subtask 3: y[i] = 0 for all i, so just do a dp where DP[i][use] = max(DP[i-2][use] + W[i], DP[i-1][!use] + w[i])
  vl Weights(N, 0LL); for (int i = 0; i < M; i++) Weights[X[i]] += W[i];
  ll tl = 0LL; for (int i = 0; i < N; i++) tl += Weights[i];
  if (N == 2) return max(Weights[0], Weights[1]);

  vl f(N+3, 0LL);
  f[N] = f[N+1] = f[N+2] = 0;
  f[N-1] = Weights[N-1];
  
  for (int i = N - 2; i >= 0; i--) {
      ll option1 = Weights[i] + f[i+2];
      ll option2 = (i+1 < N ? Weights[i+1] + f[i+3] : LLONG_MAX);
      f[i] = min(option1, option2);
  }
  
  ll bestCost = f[0];
  return tl - bestCost;
}

• Subtask 1: 3 / 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