이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define double long double
#define FOR(i,l,r,d) for(int i=(l);i<=(r);i+=(d))
#define szof(x) ((x).size())
#define vi vector<int>
#define pii pair<int,int>
#define F first
#define S second
#define pb push_back
#define eb emplace_back
#define mkp make_pair
const int INF = 1e18;
const int MOD = 1000000007;
const int mod = 998244353;
const double eps = 1e-12;
//#pragma GCC optimize("Ofast")
const int NMAX = 101;
const int MMAX = 9901;
const int CMAX = 1001;
/// c = n. of items
int n, m, c;
int buy[NMAX][CMAX], sell[NMAX][CMAX];
/// buy[i][j] : buy item j at market i
int dt[NMAX][NMAX], profit[NMAX][NMAX], weight[NMAX][NMAX];
int sp[NMAX][NMAX];
int res;
bool has_non_pos_cyc(int T){
/// check if the graph has a non-positive cycle
/// init
FOR(i,1,n,1){
FOR(j,1,n,1){
sp[i][j] = weight[i][j];
}
}
/// shortest path
FOR(k,1,n,1){
FOR(i,1,n,1){
FOR(j,1,n,1){
sp[i][j] = min(sp[i][j], sp[i][k] + sp[k][j]);
if(sp[i][j]<-INF){ /// prevent overflow
return 1;
}
}
}
}
/// check if cycle(i->j->i)<=0 for all i!=j
FOR(i,1,n,1){
FOR(j,1,n,1){
if(i!=j and sp[i][j]+sp[j][i]<=0){
return 1;
}
}
}
return 0;
}
bool check(int T){
/// check if new graph has a non-negative cycle
/// negate all edges -> check if the graph has a non-positive cycle
FOR(i,1,n,1){
FOR(j,1,n,1){
if(dt[i][j]!=INF){
weight[i][j] = -profit[i][j] + T*dt[i][j];
}
else{
weight[i][j] = INF;
}
}
}
return (has_non_pos_cyc(T));
}
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
/// input
cin>>n>>m>>c;
FOR(i,1,n,1){
FOR(j,1,c,1){
cin>>buy[i][j]>>sell[i][j];
}
}
/// dt, profit
FOR(i,1,n,1){
FOR(j,1,n,1){
dt[i][j] = INF; /// dt = INF : no edge
}
dt[i][i] = 0;
}
FOR(i,1,m,1){
int u, v, w;
cin>>u>>v>>w;
dt[u][v] = w;
}
FOR(k,1,n,1){
FOR(i,1,n,1){
FOR(j,1,n,1){
dt[i][j] = min(dt[i][j], dt[i][k] + dt[k][j]);
}
}
}
FOR(i,1,n,1){
FOR(j,1,n,1){
profit[i][j] = 0;
FOR(k,1,c,1){
if(buy[i][k]!=-1 and sell[j][k]!=-1){
profit[i][j] = max(profit[i][j], -buy[i][k] + sell[j][k]);
}
}
}
profit[i][i] = 0;
}
/// BS for the answer
int L=0, R=1e9+1, mid;
while(L<R){
mid = (L+R+1) / 2;
if(check(mid)){
L = mid;
}
else{
R = mid-1;
}
}
res = L;
cout<<res<<'\n';
return 0;
}
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