Submission #761136

#	Time	Username	Problem	Language	Result	Execution time	Memory
761136		fatemetmhr	Ideal city (IOI12_city)	C++17	55 / 100	66 ms	14744 KiB

city

//  ~ Be Name Khoda ~  //

#include <bits/stdc++.h>
//#pragma GCC optimize ("O3")
//#pragma GCC target("avx2")
//#pragma GCC optimize("unroll-loops,Ofast")

using namespace std;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second

const int maxn  =  1e6   + 10;
const int maxn5 =  5e5   + 10;
const int maxnt =  1.2e6 + 10;
const int maxn3 =  2e3   + 10;
const int mod   =  1e9;
const ll  inf   =  1e18;

int h[maxn3], q[maxn3];
vector <int> adj[maxn5];
map <pair<ll, ll>, int> p;
vector <ll> avx, avy;


inline int solve(int n, int *x, int *y){
    for(int i = 0; i < n; i++)
        p[{x[i], y[i]}] = i;
    for(int i = 0; i < n; i++){
        if(p.find(mp(x[i] - 1, y[i])) != p.end())
            adj[i].pb(p[{x[i] - 1, y[i]}]);
        if(p.find(mp(x[i] + 1, y[i])) != p.end())
            adj[i].pb(p[{x[i] + 1, y[i]}]);
        if(p.find(mp(x[i], y[i] - 1)) != p.end())
            adj[i].pb(p[{x[i], y[i] - 1}]);
        if(p.find(mp(x[i], y[i] + 1)) != p.end())
            adj[i].pb(p[{x[i], y[i] + 1}]);
    }
    ll ans = 0;
    for(int i = 0; i < n; i++){
        memset(h, -1, sizeof h);
        int l = 0, r = 1;
        q[0] = i;
        h[i] = 0;
        while(l < r){
            int v = q[l++];
            ans += h[v];
            for(auto u : adj[v]) if(h[u] == -1){
                h[u] = h[v] + 1;
                q[r++] = u;
            }
        }
    }
    ans /= 2;
    return ans % mod;
}


int DistanceSum(int n, int *x, int *y){

    if(n <= 2000)
        return solve(n, x, y);

    for(int i = 0; i < n; i++){
        avx.pb(x[i]);
        avy.pb(y[i]);
    }
    sort(all(avx));
    sort(all(avy));
    ll psx = avx[0], psy = avy[0];
    ll ans = 0;
    for(ll i = 1; i < n; i++){
        ans += i * avx[i] - psx; ans %= mod;
        ans += i * avy[i] - psy; ans %= mod;
        psx += avx[i];
        psy += avy[i];
    }

    return ans % mod;

}

• Subtask 1: 11 / 11
• Subtask 2: 21 / 21
• Subtask 3: 23 / 23
• Subtask 4: 0 / 45