이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/* made by amunduzbaev */
#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;
#define ff first
#define vv vector
#define ss second
#define pb push_back
#define mp make_pair
#define ub upper_bound
#define lb lower_bound
#define sz(x) (int)x.size()
#define tut(x) array<int, x>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(),x.rend()
#define NeedForSpeed ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define int long long
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vii;
typedef vector<pii> vpii;
template<class T> bool umin(T& a, const T& b) { return a > b? a = b, true:false; }
template<class T> bool umax(T& a, const T& b) { return a < b? a = b, true:false; }
//void usaco(string s) { freopen((s+".in").c_str(),"r",stdin);
//freopen((s+".out").c_str(),"w",stdout); }
//template<class T> tree<T,
//less<T>, null_type, rb_tree_tag,
//tree_order_statistics_node_update> ordered_set;
const int N = 105;
const int mod = 1e9+7;
const ll inf = 1e16;
const ld Pi = acos(-1);
#define MULTI 0
int n, m, k, ans, q, res, a[N];
int b[N][1005], s[N][1005];
int tt[N][N], prof[N][N];
int mx[N][N];
bool check(int m){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
mx[i][j] = -inf;
if(tt[i][j] == inf || i == j) continue;
mx[i][j] = prof[i][j] - m * tt[i][j];
}
}
for(int k=1;k<=n;k++){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
umax(mx[i][j], mx[i][k] + mx[k][j]);
}
}
}
for(int i=1;i<=n;i++) if(mx[i][i] >= 0) return 1;
return 0;
}
void solve(int t_case){
cin>>n>>m>>k;
for(int i=1;i<=n;i++){
for(int j=0;j<k;j++){
cin>>b[i][j]>>s[i][j];
}
} for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) tt[i][j] = inf;
for(int i=0;i<m;i++){
int a, b, c; cin>>a>>b>>c;
umin(tt[a][b], c);
} int K = k;
for(int k=1;k<=n;k++){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
umin(tt[i][j], tt[i][k] + tt[k][j]);
}
for(int j=0;j<K;j++){
if(~b[i][j] && ~s[k][j])
umax(prof[i][k], - b[i][j] + s[k][j]);
}
}
}
int l = 0, r = mod;
while(l < r){
int m = (l + r + 1)>>1;
if(check(m)) l = m;
else r = m-1;
} cout<<l<<"\n";
}
signed main(){
NeedForSpeed
if(!MULTI) {
solve(1);
} else {
int t; cin>>t;
for(int t_case = 1; t_case <= t; t_case++) solve(t_case);
} return 0;
}
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