Submission #922466

#	Time	Username	Problem	Language	Result	Execution time	Memory
922466		Ludissey	Catfish Farm (IOI22_fish)	C++17	0 / 100	835 ms	2097152 KiB

fish

#include "fish.h"
#include <bits/stdc++.h>
#define int long long
using namespace std;
vector<vector<int>> a;
vector<vector<int>> fsh;
vector<vector<int>> prfx;
vector<vector<int>> possprfx;
 
vector<vector<vector<int>>> dp;
int N,M;
 
long long max_weights(signed n, signed m, std::vector<signed> X, std::vector<signed> Y, std::vector<signed> W) {
  N=n; M=m; 
  fsh.resize(N+1);
  prfx.resize(N+1, vector<int>(N+1,0));
  possprfx.resize(N+1, vector<int>(N+1,0));
  dp.resize(N+1, vector<vector<int>>((N+1), vector<int>(3,0)));
  a.resize(N+1, vector<int>((N+1), 0));
  for (int i = 0; i < M; i++){
    a[X[i]][Y[i]] = W[i];
    // fsh[X[i]+1].push_back(Y[i]+1);
    // if(X[i]>0) fsh[X[i]-1].push_back(Y[i]+1);
  }
  for (int i = 0; i < N; i++) sort(fsh[i].begin(),fsh[i].end()); // N log  N
  for (int i = 0; i < N; i++) //N*N
  {
    for (int j = 0; j < N; j++)
    {
      prfx[i][j]=a[i][j];
      if(j>0) prfx[i][j]+=prfx[i][j-1];
    }
  }
  int mx=0;
  for (int x = N-1; x >= 0; x--) // O(N)
  {
    int _mx=-1, _mxI=0;
    for (int i = 0; i < N; i++)
    {
      int cmx=max(dp[x+1][i][0],max(dp[x+1][i][1],dp[x+1][i][2]));
      if(cmx>_mx) _mxI=i;
      _mx=max(_mx,cmx);
    }
    fsh[x].push_back(_mxI);
    _mx=-1, _mxI=N;
    for (int i = N; i > 0; i--)
    {
      int cmx=max(dp[x+1][i][0],max(dp[x+1][i][1],dp[x+1][i][2]));
      if(cmx>_mx) _mxI=i;
      _mx=max(_mx,cmx);
    }
    fsh[x].push_back(_mxI);

    for (int last = 0; last <= N; last++) // O(N)
    {
      if(x==0&&last>0) break;
      for (int big = 0; big <= 2; big++)// O(2)
      {
        for (auto i:fsh[x]) // O(3)
        {
          if(i>last&&big==0) continue;
          int c=0;
          if(i==0) {
            if(big!=2) c=dp[x+1][last][1+(last>0)];
            else c=dp[x+1][0][1];
          }
          else c+=dp[x+1][i][i>last];

          if(i==0&&last>0) {
            if(big<2) c+=prfx[x][last-1];
          }
          else if(last>i&&big<2) {
            c+=prfx[x][last-1]-prfx[x][i-1];
          }
          else if(i>last&&x>0){
            int ng=0;
            if(last>0) ng=prfx[x-1][last-1];
            c+=prfx[x-1][i-1]-ng;
          }
          dp[x][last][big]=max(c, dp[x][last][big]);
          mx=max(dp[x][last][big],mx);
        }
        //zcerr<<x << " " << last << " " << big << " = " << dp[x][last][big] <<  "\n";
      }
    }
  }
  return mx;
}

• 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