# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
57415 | Benq | Luxury burrow (IZhO13_burrow) | C++14 | 2055 ms | 18684 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
template<int SZ> struct DSU {
int par[SZ], sz[SZ];
DSU() {
F0R(i,SZ) par[i] = -1, sz[i] = 1;
}
int get(int x) { // path compression
if (par[x] != x) par[x] = get(par[x]);
return par[x];
}
bool unite(int x, int y) { // union-by-rank
x = get(x), y = get(y);
if (x == y) return 0;
if (sz[x] < sz[y]) swap(x,y);
sz[x] += sz[y], par[y] = x;
return 1;
}
};
int n,m,k,cur[1000];
int g[1000][1000];
ll ans = 0;
void solve(int x) {
vpi v;
F0R(i,n) v.pb({cur[i]-x,i});
sort(v.rbegin(),v.rend());
DSU<1000> D = DSU<1000>();
for (auto a: v) {
D.par[a.s] = a.s;
if (a.s > 0 && D.par[a.s-1] != -1) D.unite(a.s,a.s-1);
if (a.s < n-1 && D.par[a.s+1] != -1) D.unite(a.s,a.s+1);
ans = max(ans,a.f*(ll)D.sz[D.get(a.s)]);
}
}
int get(int lo) {
ans = 0;
F0R(i,n) cur[i] = m;
F0Rd(j,m) {
F0R(i,n) if (g[i][j] < lo) cur[i] = j;
solve(j);
}
return ans;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> n >> m >> k;
F0R(i,n) F0R(j,m) cin >> g[i][j];
int lo = 1, hi = 1000000000;
while (lo < hi) {
int mid = (lo+hi+1)/2;
if (get(mid) >= k) lo = mid;
else hi = mid-1;
}
cout << lo << " " << get(lo);
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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