| # | Time | Username | Problem | Language | Result | Execution time | Memory | 
|---|---|---|---|---|---|---|---|
| 258773 | Kastanda | Hamburg Steak (JOI20_hamburg) | C++11 | 3093 ms | 6620 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.
// M
#include<bits/stdc++.h>
using namespace std;
const int N = 200005;
struct Data {int a, b, c, d;};
int n, k, M[N];;
Data A[N];
mt19937 Rnd(time(0));
inline Data Merge(Data X, Data Y)
{
        Data R;
        R.a = max(X.a, Y.a);
        R.b = max(X.b, Y.b);
        R.c = min(X.c, Y.c);
        R.d = min(X.d, Y.d);
        return R;
}
inline bool OK(Data X)
{
        return (X.a <= X.c && X.b <= X.d);
}
int main()
{
        scanf("%d%d", &n, &k);
        for (int i = 1; i <= n; i ++)
                scanf("%d%d%d%d", &A[i].a, &A[i].b, &A[i].c, &A[i].d);
        vector < int > Priority;
        while (true)
        {
                // May God help us ...
                vector < int > P;
                memset(M, 0, sizeof(M));
                for (int i : Priority)
                        M[i] = 1, P.push_back(i);
                shuffle(P.begin(), P.end(), Rnd);
                for (int i = 1; i <= n; i ++)
                        if (!M[i]) P.push_back(i);
                shuffle(P.begin() + (int)Priority.size(), P.end(), Rnd);
                Priority.clear();
                Data F[4];
                for (int j = 0; j < k; j ++)
                        F[j].a = F[j].b = 0, F[j].c = F[j].d = 1e9 + 9;
                for (int i : P)
                {
                        bool Fail = 1;
                        for (int j = 0; j < k && Fail; j ++)
                                if (OK(Merge(A[i], F[j])))
                                        F[j] = Merge(A[i], F[j]), Fail = 0;
                        if (Fail)
                                Priority.push_back(i);
                }
                if (!Priority.size())
                {
                        for (int j = 0; j < k; j ++)
                                printf("%d %d\n", F[j].a, F[j].b);
                        return 0;
                }
        }
}
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