# Replication-USACO Gold December 2020

I was trying to solve this problem. My idea is identical to the one mentioned in the editorial (at least I think so). I basically check for each cell whether it could be a potential center of a swarm of cows. Then, having calculated every center and how many times it could replicate, I want to then check which other cells could be part of a swarm. This is the case when its manhattan distance to a center is less than or equal than the maximum number of times this center can replicate. However, I couldn´t implement this last part, and furthermore couldn’t understand what the editorial does to count this cells.
My code so far:

Summary
``````#include <bits/stdc++.h>

using namespace std;

using tint = long long;
using ld = long double;

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

using pi = pair<tint,int>;
using pl = pair<tint,tint>;
using vi = vector<int>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vvi = vector<vi>;
using vl = vector<tint>;
using vb = vector<bool>;

#define pb push_back
#define pf push_front
#define rsz resize
#define all(x) begin(x), end(x)
#define rall(x) x.rbegin(), x.rend()
#define sz(x) (int)(x).size()
#define ins insert

#define f first
#define s second
#define mp make_pair

#define DBG(x) cerr << #x << " = " << x << endl;

const int MOD = 1e9+7;
const tint mod = 998244353;
const int MX = 1005;
const tint INF = 1e18;
const int inf = 2e9;
const ld PI = acos(ld(-1));
const ld eps = 1e-8;

const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};

template<class T> void remDup(vector<T> &v){
sort(all(v)); v.erase(unique(all(v)),end(v));
}

template<class T> bool ckmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0;
}
template<class T> bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0;
}

bool valid(int x, int y, int n, int m){
return (0<=x && x<n && 0<=y && y<m);
}

tint cdiv(tint a, tint b) { return a/b+((a^b)>0&&a%b); } //redondea p arriba
tint fdiv(int a, int b) { return a/b-((a^b)<0&&a%b); } //redondea p abajo

void NACHO(string name = ""){
ios_base::sync_with_stdio(0); cin.tie(0);
if(sz(name)){
freopen((name+".in").c_str(), "r", stdin);
freopen((name+".out").c_str(), "w", stdout);
}
}

string a[MX];
int distr[MX][MX], dists[MX][MX], distc[MX][MX];
bool visited[MX][MX], center[MX][MX], stop[MX][MX];

int main(){
NACHO();
int n, d; cin >> n >> d;
F0R(i, n) cin >> a[i];
queue<pi> q;
F0R(i, n){
F0R(j, n){
if(a[i][j] == '#'){
q.push(mp(i, j));
visited[i][j] = 1;
}
}
}

// cual es la piedra mas cercana.
while(sz(q)){
auto u = q.front();
int x = u.f, y = u.s;
q.pop();
F0R(k, 4){
int xx = x+dx[k], yy = y+dy[k];
if(valid(xx, yy, n, n) && !visited[xx][yy]){
visited[xx][yy] = 1;
distr[xx][yy] = distr[x][y]+1;
q.push(mp(xx, yy));
}
}
}

while(sz(q)) q.pop();
F0R(i, n){
F0R(j, n){
visited[i][j] = 0;
if(a[i][j] == 'S'){
q.push(mp(i, j));
visited[i][j] = 1;
}
}
}

// calculo la distancia desde cada celda
// a la celda de comienzo mas cercana.
// de esta forma puedo saber si una celda
// es un potencial centro de rebaño
// ya que si distcomienzo/d < distpiedra
// se que puede ser centro
while(sz(q)){
auto u = q.front();
int x = u.f, y = u.s;
q.pop();
center[x][y] = 1;
if(dists[x][y]/d >= distr[x][y]) continue;
F0R(k, 4){
int xx = x+dx[k], yy = y+dy[k];
if(valid(xx, yy, n, n) && !visited[xx][yy] && a[xx][yy] != '#'){
visited[xx][yy] = 1;
dists[xx][yy] = dists[x][y]+1;
q.push(mp(xx, yy));
}
}
}

// ahora, veo que celdas son parte de la replicacion
F0R(i, n){
F0R(j, n) cout << center[i][j];
cout << endl;
}

}
``````

The array “stop” indicates whether a certain cell can be a center, but cannot replicate right after.
Hope I made my explanation clear enough.

ew look at all those macros
Ok, but this does involve 3 BFS’s, and you got the first 2 first. The third one needs some clever thinking. You add the new centers which are closer to rocks gradually and expand the rest as you go. You can still keep a `visited` array with this BFS, so the complexity is still O(N^2).