Submission #1040079

#	Time	Username	Problem	Language	Result	Execution time	Memory
1040079		ProtonDecay314	Friend (IOI14_friend)	C++17	0 / 100	1 ms	448 KiB

friend

#include "friend.h"
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
typedef vector<vpi> vvpi;
typedef vector<vpll> vvpll;
typedef vector<bool> vb;
#define IOS ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#define L(varll, mn, mx) for(ll varll = (mn); varll < (mx); varll++)
#define LR(varll, mx, mn) for(ll varll = (mx); varll > (mn); varll--)
#define LI(vari, mn, mx) for(int vari = (mn); vari < (mx); vari++)
#define LIR(vari, mx, mn) for(int vari = (mx); vari > (mn); vari--)
#define INPV(varvec) for(auto& varveci : (varvec)) cin >> varveci
#define fi first
#define se second
#define pb push_back
#define INF(type) numeric_limits<type>::max()
#define NINF(type) numeric_limits<type>::min()
#define TCASES int t; cin >> t; while(t--)

// Find out best sample
int findSample(int n,int confidence[],int host[],int protocol[]){
    vvi adj;
    for(int i = 0; i < n; i++) {
        vi adjr;
        adj.pb(adjr);
    }

    for(int i = 1; i < n; i++) {
        if(protocol[i] == 0) {
            // I am your friend
            adj[i].pb(host[i]);
            adj[host[i]].pb(i);
        } else if(protocol[i] == 1) {
            // My friends are your friends
            for(int j : adj[host[i]]) {
                adj[i].pb(j);
                adj[j].pb(i);
            }
        }
    }

    vb vis(n, false);

    queue<pi> q;

    q.push({0, 0});

    int num0 = 0;

    while(!q.empty()) {
        auto [i, c] = q.front();
        q.pop();

        if(vis[i]) continue;
        vis[i] = true;
        if(c == 0) num0++;

        for(int j : adj[i]) {
            if(vis[j]) continue;
            q.push({j, 1 - c});
        }
    }
	return max(num0, n - num0);
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 8
• Subtask 3: 0 / 8
• Subtask 4: 0 / 19
• Subtask 5: 0 / 23
• Subtask 6: 0 / 31