Submission #1079619

#	Time	Username	Problem	Language	Result	Execution time	Memory
1079619		c2zi6	Longest Trip (IOI23_longesttrip)	C++17	25 / 100	918 ms	1544 KiB

longesttrip

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "longesttrip.h"

int n;
VVI gp;
VVI edges;

VI rev(VI a) {
    reverse(all(a));
    return a;
}

VI longest_trip(int N, int D) {
    assert(D == 1);
    n = N;
    gp = VVI(n);
    edges = VVI(n, VI(n));
    replr(i, 0, n-1) {
        replr(j, i+1, n-1) {
            if (are_connected({i}, {j})) {
                gp[i].pb(j);
                gp[j].pb(i);
                edges[i][j] = edges[j][i] = true;
            }
        }
    }
    VI path1;
    VI path2;
    path1.pb(0);
    path2.pb(1);
    replr(u, 2, n-1) {
        if (edges[path1.back()][u]) path1.pb(u);
        else if (edges[path2.back()][u]) path2.pb(u);
        else {
            for (int u : rev(path2)) path1.pb(u);
            path2 = VI{u};
        }
    }

    if (edges[path2[0]][path1[0]]) {
        VI ans;
        for (int x : rev(path2)) ans.pb(x);
        for (int x : path1) ans.pb(x);
        return ans;
    }
    if (edges[path2.back()][path1[0]]) {
        VI ans;
        for (int x : path2) ans.pb(x);
        for (int x : path1) ans.pb(x);
        return ans;
    }
    /* now I know that path2 is also a cycle */
    if (edges[path1.back()][path2[0]]) {
        VI ans;
        for (int x : path1) ans.pb(x);
        for (int x : path2) ans.pb(x);
        return ans;
    }
    /* now I know that path2 is also a cycle */



    if (path1.size() > path2.size()) return path1;
    return path2;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 5
• Subtask 2: 0 / 10
• Subtask 3: 25 / 25
• Subtask 4: 0 / 60