Submission #57416

#	Time	Username	Problem	Language	Result	Execution time	Memory
57416		Benq	Luxury burrow (IZhO13_burrow)	C++14	0 / 100	2 ms	376 KiB

burrow

#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;
vi nex[1000];

void solve(int x) {
    F0R(i,m) nex[i].clear();
    F0R(i,n) nex[cur[i]-x].pb(i);
    
    DSU<1000> D = DSU<1000>();
    F0Rd(i,m) for (int a: nex[i]) {
        D.par[a] = a;
        if (a > 0 && D.par[a-1] != -1) D.unite(a,a-1);
        if (a < n-1 && D.par[a+1] != -1) D.unite(a,a+1);
        ans = max(ans,i*(ll)D.sz[D.get(a)]);
    }
}

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
*/

• Subtask 1: 0 / 100