176A - Trading Business - CodeForces Solution


greedy sortings *1200

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C++ Code:

#include<bits/stdc++.h>

#define ll long long
#define vc vector
#define trav(a,x) for(auto& a : x)
#define all(v) v.begin(), v.end()
#define f(p,q,r) for(ll p=q;p<r;p++)
#define fr(p,q,r) for(ll p=q;p>=r;p--)
#define sz(x) (ll)(x).size()
#define el endl
#define pb push_back
#define pp pop_back
#define max_3(a,b,c) max(a,max(b,c))
#define min_3(a,b,c) min(a,min(b,c))
#define max_4(a,b,c,d) max(a,max(b,max(c,d)))
#define min_4(a,b,c,d) min(a,min(b,min(c,d)))
#define better_one ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
using namespace std;

const int N=105;
int a[N][N],b[N][N],c[N][N];
void solve()
{
    int n,m,k;
    cin >> n >> m >> k;
    string s;
    f(i,1,n+1){
        cin >> s;
        f(j,1,m+1) cin >> a[i][j] >> b[i][j] >> c[i][j];
    }
    auto t=[&](int x,int y)
    {
        bool used[m+1]={0};
        int ans=0,last=k;
        while(last)
        {
            int mx=0,ix,ct;
            f(i,1,m+1){
                if(used[i]) continue;
                if(b[y][i]-a[x][i]>mx)
                {
                    ix=i;
                    mx=b[y][i]-a[x][i];
                }
            }
            if(mx==0) break;
            ct=min(last,c[x][ix]);
            ans+=ct*mx;
            used[ix]=1;
            last-=ct;
        }
        return ans;
    };
    int ans=0;
    f(i,1,n+1)
    {
        f(j,1,n+1) ans=max(ans,t(i,j));
    }
    cout<<ans;
}

int main ()
{
    cout << el;
    better_one;
    int t=1;
    // cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}


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