Submission #1075272

#	Time	Username	Problem	Language	Result	Execution time	Memory
1075272		mariaclara	Catfish Farm (IOI22_fish)	C++17	100 / 100	127 ms	37972 KiB

fish

#include "fish.h"
#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef pair<ll,ll> pii;
typedef tuple<int,int,int> trio;
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define pb push_back
#define mk make_pair
#define fr first 
#define sc second
 
ll max_weights(int N, int M, vector<int> X, vector<int> Y, vector<int> W) {
    vector<vector<pii>> val(N);
    vector<vector<ll>> dp[2];
    dp[0].resize(N); // x-1 < x
    dp[1].resize(N); // x-1 > x
 
    for(int i = 0; i < M; i++) val[X[i]].pb({Y[i], W[i]});
    
    dp[0][0].resize(sz(val[0])+1, 0);
    dp[1][0].resize(sz(val[0])+1, 0);
    sort(all(val[0]));
    reverse(all(val[0]));
 
    for(int x = 1; x < N; x++) {
        sort(all(val[x]));
        reverse(all(val[x]));
 
        dp[0][x].resize(sz(val[x])+1);
        dp[1][x].resize(sz(val[x])+1);
 
        // >
        for(int i = 0, j = 0; i < sz(val[x]); i++) {
            auto [y,w] = val[x][i];
            if(i != 0) dp[1][x][i+1] = dp[1][x][i] + w;
 
            while(j < sz(val[x-1]) and val[x-1][j].fr > y) j++;
            
            dp[1][x][i+1] = max(dp[1][x][i+1], dp[1][x-1][j] + w); // LEMBRAR DE dp[1][x][j] max= dp[1][x][j-1]
            //cout << "1x = " << x << ", i = " << i << ", y = " << y << ", w = " << w << ", dp = " << dp[1][x][i+1] << "\n";
        }
 
        // <
        ll w = 0;
        for(int i = sz(val[x]) - 1, j = sz(val[x-1]) - 1, y; i >= -1; i--) {
            if(i >= 0) y = val[x][i].fr;
            else y = N;
  
            while(j >= 0 and val[x-1][j].fr < y) {
                w = max(w, dp[0][x-1][j+1]);
                w += val[x-1][j].sc;
                j--;
            }
 
            dp[0][x][i+1] = max({dp[0][x][i+1], dp[1][x-1][sz(val[x-1])], w});
            //cout << "0x = " << x << ", i = " << i << ", y = " << y << ", w = " << w << ", dp = " << dp[0][x][i+1] << "\n";
        }
 
        dp[1][x][0] = max(dp[1][x][0], dp[0][x][0]);
        for(int i = 1; i <= sz(val[x]); i++)
            dp[1][x][i] = max({dp[1][x][i], dp[1][x][i-1], dp[0][x][i]});
    }
 
    return dp[1][N-1][sz(val[N-1])];
}

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