#include "longesttrip.h"
using namespace std;
#include <bits/stdc++.h>
#define F(i, l, r) for (int i = (l); i < (r); ++i)
#define A(a) (a).begin(), (a).end()
int edge[300][300];
int parent[300];
int find(int i) {
return i == parent[i] ? i : parent[i] = find(parent[i]);
}
void add_edge(int A, int B) {
edge[A][B] = edge[B][A] = 1;
parent[find(A)] = find(B);
}
bool query(vector<int> A, vector<int> B) {
// for (auto x: A ) cout << x << " "; cout << endl;
// for (auto x: B ) cout << x << " "; cout << endl;
if (A.size() == 1 and B.size() == 1) {
if (edge[A[0]][B[0]]) return 1;
}
bool res = are_connected(A, B);
if (res) {
if (A.size() == 1 and B.size() == 1) {
add_edge(A[0], B[0]);
}
}
return res;
}
vector<int> longest_trip(int N, int D)
{
iota(parent, parent + N, 0ll);
memset(edge, 0, sizeof edge);
deque<int> hamil;
auto gen = [&]() {
set<int> v;
vector<bool> seen(N+1);
for (auto x: hamil) seen[x] = 1;
F(i, 0, N) if (!seen[i]) v.insert(i);
return v;
};
if (query({0}, {1})) {
hamil = {0, 1};
} else if (query({1}, {2})) {
hamil = {1, 2};
} else {
add_edge(0, 2);
hamil = {0, 2};
}
while (hamil.size() != N) { // this will reach N, with the exception of one case
for (auto x: hamil) cout << x << " "; cout << endl;;
// Mandatory 1 query here.
// If we go into other branch, good, we only need 2 queries at most.
if (query({hamil[0]}, {hamil.back()})) {
// behaves as hamil cycle
} else {
// Pick any node not in hamil, this must be connected to either the front or the back.
auto i = *gen().begin();
// cout << "HAHA " << endl;
if (query({hamil[0]}, {i})) hamil.push_front(i);
else {
add_edge(hamil.back(), i);
hamil.push_back(i);
}
continue;
}
// Okay, we want to find *any* connection between hamil and others, if exists already.
auto bad = gen();
bool seen_edge = false;
auto prev = hamil;
F(i, 0, hamil.size()) {
auto v = hamil.back(); hamil.pop_back(); hamil.push_front(v);
for (auto y: bad) if (edge[v][y]) {
hamil.push_front(y);
seen_edge = 1;
goto end;
}
}
end: if (seen_edge) continue;
assert(prev == hamil);
// cout << " ivaldf " << endl;
// Okay, now we want to do the "pick one from end, and two disjoint components" strategy.
while (true) {
int state = 0;
// Amortized shit;
// OOOH. But here it *can* overlap, we pay 1 mandatory per thing and 2 mandatory per amortized, so in total
// 3N queries worst case... hmm...
for (auto x: bad) for (auto y: bad) if (find(x) != find(y)) {
// cout << "YA " << x << " " << y << endl;
if (query({x}, {y})) {
// cout << "Shitty " << endl;
state = 1; goto end2; // incr spanning tree, okay.
} else { // actually no edge between these, therefore one end MUST connect to hamil[0].
state = 2;
// cout << " HOW THO " << x << " " << y << " " << hamil[0] << endl;
if (query({x}, {hamil[0]})) {
hamil.push_front(x);
} else {
add_edge(hamil[0], y);
hamil.push_front(y);
}
goto end2;
}
}
end2:
if (state == 1) continue;
if (state == 2) break;
// Mandatory pay once, 2 * log(n) queries
// OTHERWISE, RARE ONCE IN AN ITERATION CASE: WE ASK IF THERE'S ANY CONNECTION ACROSS BOTH COMPS.
// IF NOT, THEN INSTANTLY KNOW TWO K_N COMPS.
// cout << "CHECKING KN " << endl;
if (!query(vector<int>(A(hamil)), vector<int>(A(bad)))) {
if (hamil.size() < bad.size()) {
return vector<int>(A(bad));
}
return vector<int>(A(hamil));
}
// ELSE, FIND ANY SPANNING EDGE WITH BSEARCH, TAKE IT. BREAK, WE WILL FIND THE SPANNING EDGE AGAIN NEXT LOOP.
vector<int> cur(A(bad));
while (cur.size() > 1) {
vector<int> shit(cur.begin(), cur.begin() + cur.size()/2);
if (query(vector<int>(A(hamil)), shit)) cur = shit;
else cur = vector<int>(cur.begin() + cur.size() / 2, cur.end());
}
int x = cur[0];
cur = vector<int>(A(hamil));
while (cur.size() > 1) {
vector<int> shit(cur.begin(), cur.begin() + cur.size()/2);
if (query({x}, shit)) cur = shit;
else cur = vector<int>(cur.begin() + cur.size() / 2, cur.end());
}
int y = cur[0];
add_edge(x, y);
break;
}
// cout << "WHILE TRUING " << endl;
}
return vector<int>(A(hamil));
}
Compilation message
longesttrip.cpp: In function 'std::vector<int> longest_trip(int, int)':
longesttrip.cpp:63:25: warning: comparison of integer expressions of different signedness: 'std::deque<int>::size_type' {aka 'long unsigned int'} and 'int' [-Wsign-compare]
63 | while (hamil.size() != N) { // this will reach N, with the exception of one case
| ~~~~~~~~~~~~~^~~~
longesttrip.cpp:64:9: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
64 | for (auto x: hamil) cout << x << " "; cout << endl;;
| ^~~
longesttrip.cpp:64:47: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
64 | for (auto x: hamil) cout << x << " "; cout << endl;;
| ^~~~
longesttrip.cpp:5:40: warning: comparison of integer expressions of different signedness: 'int' and 'std::deque<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
5 | #define F(i, l, r) for (int i = (l); i < (r); ++i)
| ^
longesttrip.cpp:91:9: note: in expansion of macro 'F'
91 | F(i, 0, hamil.size()) {
| ^
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
0 ms |
600 KB |
Secret mismatch (possible tampering with the output). |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
1 ms |
600 KB |
Secret mismatch (possible tampering with the output). |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
1 ms |
600 KB |
Secret mismatch (possible tampering with the output). |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
0 ms |
600 KB |
Secret mismatch (possible tampering with the output). |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
결과 |
실행 시간 |
메모리 |
Grader output |
| 1 |
Incorrect |
1 ms |
600 KB |
Secret mismatch (possible tampering with the output). |
| 2 |
Halted |
0 ms |
0 KB |
- |
