Submission #78814

#	Time	Username	Problem	Language	Result	Execution time	Memory
78814		wleung_bvg	Highway Tolls (IOI18_highway)	C++14	0 / 100	52 ms	2936 KiB

highway

#include "highway.h"
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define f first
#define s second
#define all(a) (a).begin(),(a).end()
#define For(i,a,b) for(auto i=(a);i<(b);i++)
#define FOR(i,b) For(i,0,b)
#define Rev(i,a,b) for(auto i=(a);i>(b);i--)
#define REV(i,a) Rev(i,a,-1)
#define FORE(i,a) for(auto&&i:a)
#define sz(a) (int((a).size()))
#define MIN(a,b) ((a)=min((a),(b)))
#define MAX(a,b) ((a)=max((a),(b)))
using ll=long long;using ld=long double;using uint=unsigned int;using ull=unsigned long long;
using pii=pair<int,int>;using pll=pair<ll,ll>;using pill=pair<int,ll>;using plli=pair<ll,int>;using pdd=pair<double,double>;using pld=pair<ld,ld>;
constexpr const char nl='\n',sp=' ';constexpr const int INT_INF=0x3f3f3f3f;constexpr const ll LL_INF=0x3f3f3f3f3f3f3f3f;
constexpr const double D_INF=numeric_limits<double>::infinity();constexpr const ld LD_INF=numeric_limits<ld>::infinity();constexpr const double EPS=1e-9;

const int MAXN = 90005;

bool vis[MAXN];
int q[MAXN];
vector<int> adj[MAXN];

void find_pair(int N, vector<int> U, vector<int> V, int A, int B) {
    int M = sz(U);
    FOR(i, M) {
        adj[U[i]].pb(V[i]);
        adj[V[i]].pb(U[i]);
    }
    vector<int> Q(M, 0);
    ll DA = ask(Q);
    int s = 0;
    vector<int> ans;
    FOR(_, 2) {
        fill(vis, vis + N, 0);
        int front = 0, back = 0, lo = 0, hi = N - 1, mid;
        vis[q[back++] = s] = 1;
        while (front < back) FORE(w, adj[q[front++]]) if (!vis[w]) vis[q[back++] = w] = 1;
        while (lo <= hi) {
            mid = lo + (hi - lo) / 2;
            FOR(i, mid) vis[q[i]] = 0;
            For(i, mid, N) vis[q[i]] = 1;
            FOR(i, M) Q[i] = vis[U[i]] && vis[V[i]];
            if (ask(Q) == DA) hi = mid - 1;
            else lo = mid + 1;
        }
        ans.pb(s = q[hi]);
    }
    answer(ans[0], ans[1]);
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 7
• Subtask 3: 0 / 6
• Subtask 4: 0 / 33
• Subtask 5: 0 / 18
• Subtask 6: 0 / 31