This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,popcnt,sse4,abm")
#include <bits/stdc++.h>
using namespace std;
#ifndef WAIMAI
#include "longesttrip.h"
#else
#include "grader.cpp"
#endif
#ifdef WAIMAI
#define debug(HEHE...) cout << "[" << #HEHE << "] : ", dout(HEHE)
void dout() {cout << '\n';}
template<typename T, typename...U>
void dout(T t, U...u) {cout << t << (sizeof...(u) ? ", " : ""), dout(u...);}
#else
#define debug(...) 7122
#endif
#define ll long long
#define Waimai ios::sync_with_stdio(false), cin.tie(0)
#define FOR(x,a,b) for (int x = a, I = b; x <= I; x++)
#define pb emplace_back
#define F first
#define S second
bool is(int x, int y) {
return are_connected({x}, {y});
}
vector<int> longest_trip(int n, int D) {
vector<int> A, B;
A.pb(0), B.pb(1);
FOR (i, 2, n - 1) {
if (is(A.back(), i)) {
A.pb(i);
continue;
}
if (is(B.back(), i)) {
B.pb(i);
continue;
}
reverse(A.begin(), A.end());
A.insert(A.end(), B.begin(), B.end());
B.clear();
B.pb(i);
}
if (is(A[0], B[0])) {
reverse(A.begin(), A.end());
A.insert(A.end(), B.begin(), B.end());
return A;
}
if (is(A[0], B.back())) {
B.insert(B.end(), A.begin(), A.end());
return B;
}
if (is(A.back(), B[0])) {
A.insert(A.end(), B.begin(), B.end());
return A;
}
if (is(A.back(), B.back())) {
reverse(B.begin(), B.end());
A.insert(A.end(), B.begin(), B.end());
return A;
}
if (are_connected(A, B) == 0) return A.size() > B.size() ? A : B;
int i = -1, j = -1;
{
int l = 0, r = A.size() - 1;
while (l < r) {
int mid = (l + r) / 2;
vector<int> Al(A.begin() + l, A.begin() + mid + 1);
if (are_connected(Al, B)) r = mid;
else l = mid + 1;
}
i = l;
}
{
int l = 0, r = B.size() - 1;
while (l < r) {
int mid = (l + r) / 2;
vector<int> Bl(B.begin() + l, B.begin() + mid + 1);
if (are_connected({A[i]}, B)) r = mid;
else l = mid + 1;
}
j = l;
}
vector<int> ans;
FOR (k, i + 1, A.size() - 1) ans.pb(A[k]);
FOR (k, 0, i) ans.pb(A[k]);
FOR (k, j, B.size() - 1) ans.pb(B[k]);
FOR (k, 0, j - 1) ans.pb(B[k]);
return ans;
}
/*
in1
2
5 1
1
1 1
0 0 1
0 0 0 1
4 1
1
0 0
0 0 1
out1
5
1 0 2 3 4
4
2
0 1
3
4
in2
out2
*/
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