Submission #583591

#	Time	Username	Problem	Language	Result	Execution time	Memory
583591		BhavayGoyal	Cloud Computing (CEOI18_clo)	C++14	0 / 100	106 ms	262144 KiB

clo

#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

template<class T> using oset = 
            tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define ll long long
#define ld long double
#define ar array
#define vi vector<int>
#define vii vector<vector<int>>
#define pii pair<int, int>
#define pb push_back
#define all(x) x.begin(), x.end()
#define f first
#define s second

const int MOD = 1e9+7;
const int inf = 1e9;
const ll linf = 1e18;

const int d4i[4]={-1, 0, 1, 0}, d4j[4]={0, 1, 0, -1};
const int d8i[8]={-1, -1, 0, 1, 1, 1, 0, -1}, d8j[8]={0, 1, 1, 1, 0, -1, -1, -1};


// -------------------------------------------------- Main Code --------------------------------------------------

int n, m;
array<int, 3> arr[4000];
ll dp[2001][2000*50+1];

ll solve(int idx, int numOfProc) {
    if (idx > n+m || numOfProc < 0) return -linf;
    if (idx == n+m) return 0;
    if (dp[idx][numOfProc] != -1) return dp[idx][numOfProc];
    ll ans = 0;
    ans = max(ans, solve(idx+1, numOfProc));
    ans = max(ans, solve(idx+1, numOfProc+arr[idx][0]) + arr[idx][2]);
    return dp[idx][numOfProc] = ans;
}

void sol() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> arr[i][0] >> arr[i][1] >> arr[i][2];
        arr[i][2] *= -1;
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
        cin >> arr[i+n][0] >> arr[i+n][1] >> arr[i+n][2];
        arr[i+n][0] *= -1;
    }
    sort (arr, arr+n+m, [](array<int, 3> a, array<int, 3> b) {return a[1] > b[1];});
    memset(dp, -1, sizeof dp);
    cout << solve(0, 0) << endl;
}

int main () {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    // cin >> t; 
    while (t--) {
        sol();
    }
    return 0;
}

• Subtask 1: 0 / 18
• Subtask 2: 0 / 18
• Subtask 3: 0 / 18
• Subtask 4: 0 / 18
• Subtask 5: 0 / 18
• Subtask 6: 0 / 10