Submission #767329

#	Time	Username	Problem	Language	Result	Execution time	Memory
767329		GrindMachine	Travelling Merchant (APIO17_merchant)	C++17	100 / 100	72 ms	2236 KiB

merchant

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://usaco.guide/problems/apio-2017traveling-merchant/solution
https://pastebin.com/raw/ZrNfjLpg (make sure to handle overflows!)
some submissions to identify the bug in my code (found bug in taking inputs)


profit = cost/length
lets say profit = cost (we ignore length for now)

how to solve for this case?

for every pair (u,v), create an edge if v is reachable from u
this graph says: if we are @u with an empty bag, we buy something @u, then sell it @v, what's the max profit?
the weights of the edges denote this profit
(we can also choose to buy nothing, weight = 0)

find the max cost cycle (inf possible)

come back to original problem

profit = cost/length
we also consider length

so when buying something from u and selling it @v, it's optimal to take the shortest path from u to v
so for every pair (u,v), we compute:
(max_add, shortest_path)

profit function is a ratio, annoying to deal with

think b.s in such cases

how to check if profit >= mid?
cost/length >= mid
cost >= mid*length
cost - mid*length >= 0

edge (max_add, shortest_path) becomes:
edge with weight (max_add - mid*shortest_path)

find if there is a cycle with sum of weights >= 0
=> floyd-warshalls

*/

const int MOD = 1e9 + 7;
const int N = 100 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
const int K = 1e3 + 5;

ll buy[N][K], sell[N][K];
ll sp[N][N], best_buy_sell[N][N];
ll weight[N][N], lp[N][N]; // lp = longest path

void solve(int test_case)
{
    ll n,m,k; cin >> n >> m >> k;
    rep1(i,n){
        rep1(j,k){
            cin >> buy[i][j] >> sell[i][j];
        }
    }

    memset(sp,0x10,sizeof sp);

    rep1(i,m){
        ll u,v,w; cin >> u >> v >> w;
        sp[u][v] = w;
    }

    rep1(p,n){
        rep1(i,n){
            rep1(j,n){
                amin(sp[i][j], sp[i][p] + sp[p][j]);
            }
        }
    }

    rep1(i,n){
        rep1(j,n){
            if(i == j) conts;
            if(sp[i][j] >= inf2) conts;

            rep1(p,k){
                if(buy[i][p] != -1 and sell[j][p] != -1){
                    ll val = sell[j][p] - buy[i][p];
                    amax(best_buy_sell[i][j], val);
                }
            }
        }
    }

    auto ok = [&](ll mid){
        memset(weight,-0x10,sizeof weight);
        memset(lp,-0x10,sizeof lp);

        rep1(i,n){
            rep1(j,n){
                if(i == j) conts;
                if(sp[i][j] >= inf2) conts;
                ll w = best_buy_sell[i][j] - mid*sp[i][j];
                if(w > -inf2){
                    weight[i][j] = lp[i][j] = w;                                
                }
            }
        }

        rep1(p,n){
            rep1(i,n){
                rep1(j,n){
                    amax(lp[i][j], lp[i][p] + lp[p][j]);
                    amin(lp[i][j], inf2);
                }
            }
        }

        rep1(i,n){
            rep1(j,n){
                if(i == j) conts;

                ll w = lp[i][j] + lp[j][i];
                if(w >= 0){
                    return true;
                }
            }
        }

        return false;
    };

    ll l = 0, r = inf1;
    ll ans = -inf2;

    while(l <= r){
        ll mid = (l+r) >> 1;
        if(ok(mid)){
            ans = mid;
            l = mid+1;
        }
        else{
            r = mid-1;
        }
    }

    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 12 / 12
• Subtask 2: 21 / 21
• Subtask 3: 33 / 33
• Subtask 4: 34 / 34