제출 #670580

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
670580		Payam	Odašiljači (COCI20_odasiljaci)	C++17	70 / 70	147 ms	4460 KiB

odasiljaci

#include<bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define forp(i,l,r) for (int i = (l); i < (r); i ++)
#define forp1(j,l,r) for (int j = (l); j < (r); j ++)
#define form(i,l,r) for (int i = (r); i >= (l); i --)
#define form1(j,l,r) for (int j = (r); j >= (l); j --)
#define minset(x) (*((x).begin()))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define setp cout << fixed << setprecision(15)
#define all(x) (x).begin(), (x).end()
#define IOS ios::sync_with_stdio(false)
typedef long long ll;

using namespace std;
const int MAX_N = 1e3 + 13, MOD = 1e9 + 7, inf = 1e9 + 19;
ll INF = 1e18 + 118;
vector<int> adj[MAX_N];
pii cor[MAX_N];
bool vis[MAX_N];

void dfs(int root)
{
    vis[root] = 1;
    for (int u: adj[root]) {
        if (!vis[u]) {
            dfs(u);
        }
    }
}

int main()
{
    IOS;
    cin.tie(0);
    int n;
    cin >> n;
    forp(i, 1, n + 1) {
        cin >> cor[i].ff >> cor[i].ss;
    }
    double l = 0, r = inf, mid, difx, dify;
    forp(i, 0, 51) {
        bool cmp = 0;
        mid = (l + r) / 2;
        forp(i, 1, n + 1) {
            forp1(j, i + 1, n + 1) {
                difx = abs(cor[i].ff - cor[j].ff);
                dify = abs(cor[i].ss - cor[j].ss);
                if (sqrt((difx * difx) + (dify * dify)) <= mid) {
                    adj[i].pb(j);
                    adj[j].pb(i);
                }
            }
        }
        forp(i, 1, n + 1) {
            if (!vis[i]) {
                if (!cmp) {
                    cmp = 1;
                    dfs(i);
                }
                else {
                    cmp = 0;
                    break;
                }
            }
        }
        forp(i, 1, n + 1) {
            adj[i].clear();
        }
        if (!cmp)
            l = mid;
        else
            r = mid;
        fill(vis, vis + n + 2, 0);
    }
    setp << r / 2 << '\n';
    return 0;
}

• Subtask 1: 70 / 70