제출 #580159

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
580159		MohammadAghil	여행하는 상인 (APIO17_merchant)	C++17	100 / 100	93 ms	4200 KiB

merchant

      #include <bits/stdc++.h>
//   #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
    using namespace std;
  typedef long long ll;
   typedef pair<int, int> pp;
    #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
      #define per(i,r,l) for(int i = (r); i >= (l); i--)
        #define rep(i,l,r) for(int i = (l); i < (r); i++)
           #define all(x) x.begin(), x.end()
              #define sz(x) (int)(x).size()
                  #define pb push_back
                      #define ss second
                           #define ff first
                                   void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 1e2 + 5, maxk = 1e3 + 5, lg = 22, inf = ll(1e18) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll);return k*k%md*(b&1ll?a:1)%md;}

ll adj[maxn][maxn], adj2[maxn][maxn], s[maxn][maxk], b[maxn][maxk], prof[maxn][maxn];
int n;

void feloyd(ll adj[maxn][maxn]){
     rep(k,0,n) rep(i,0,n) rep(j,0,n) adj[i][j] = min(adj[i][j], adj[i][k] + adj[k][j]);
}

int main(){
     cin.tie(0) -> sync_with_stdio(0);
     int m, k; cin >> n >> m >> k;
     rep(i,0,n){
          rep(j,0,k) cin >> b[i][j] >> s[i][j];
     }
     rep(i,0,n) rep(j,0,n) adj[i][j] = inf;
     rep(i,0,m){
          int u, v, w; cin >> u >> v >> w; u--, v--;
          adj[u][v] = w;
     }
     feloyd(adj);
     rep(i,0,n) rep(j,0,n) rep(t,0,k) if(s[j][t] + 1 && b[i][t] + 1) prof[i][j] = max(prof[i][j], s[j][t] - b[i][t]);
     ll l = 0, r = ll(1e9) + 5;
     while(l + 1 < r){
          ll mid = (l + r)>>1ll;
          rep(i,0,n) rep(j,0,n) {
               if(adj[i][j] == inf) adj2[i][j] = inf;
               else adj2[i][j] = mid*adj[i][j] - prof[i][j];
          }
          feloyd(adj2);
          bool ok = false;
          rep(i,0,n) if(adj2[i][i] <= 0) ok = true;
          (ok?l:r) = mid; 
     }
     cout << l << '\n';
     return 0; 
}

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