Submission #990396

#	Time	Username	Problem	Language	Result	Execution time	Memory
990396		hariaakas646	Catfish Farm (IOI22_fish)	C++17	52 / 100	923 ms	2097152 KiB

fish

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

using namespace std;

#define scd(t) scanf("%d", &t)
#define sclld(t) scanf("%lld", &t)
#define forr(i, j, k) for (int i = j; i < k; i++)
#define frange(i, j) forr(i, 0, j)
#define all(cont) cont.begin(), cont.end()
#define mp make_pair
#define pb push_back
#define f first
#define s second
typedef long long int lli;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<lli> vll;
typedef vector<string> vs;
typedef vector<pii> vii;
typedef vector<vi> vvi;
typedef map<int, int> mpii;
typedef set<int> seti;
typedef multiset<int> mseti;
typedef long double ld;

int n, m;

long long max_weights(int N, int M, std::vector<int> X, std::vector<int> Y,
                      std::vector<int> W) {
    n = N;
    m = M;
    vector<vll> grid(n, vll(n+1));
    frange(i, m) {
        grid[X[i]][Y[i]+1] = W[i];
    }

    vector<vll> dp1(n, vll(n+1)), dp2(n, vll(n+1));
    forr(i, 1, n+1) {
        dp2[0][i] = grid[0][i];
        dp2[0][i] += dp2[0][i-1];
    }

    forr(i, 1, n) {
        lli ma = 0;
        frange(j, n+1) ma = max(ma, dp1[i-1][j]);
        deque<lli> dq;
        lli tot = 0;
        frange(j, n+1) {
            
            tot += grid[i][j];
            lli v = dp2[i-1][j] - tot;
            while(dq.size() && dq.back() <= v) dq.pop_back();
            dq.pb(v);
            dp2[i][j] = dq.front() + tot; 
            
            dp2[i][j] = max(dp2[i][j], ma + tot);
            
        }
        tot = 0;
        dq = {};
        for(int j=n; j>=0; j--) {
            dp1[i][j] = dp2[i-1][j];
            if(dq.size()) {
                dp1[i][j] = max(dp1[i][j], dq.front() + tot);
            }
            lli v = dp1[i-1][j] - tot;
            while(dq.size() && dq.back() <= v) dq.pop_back();
            dq.pb(v);
            tot += grid[i][j];
        }
    }   
    return *max_element(all(dp1[n-1]));
}

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