# Telephone USACO 2021

I managed to solve this problem with the following idea: I build a graph in which each node represents a position of the array and I create an edge between i and j with cost abs(i-j) if this connection is valid and j is the closest position with that breed to the right or to the left of i. After reading the editorial, I realised that this approach is somewhat similar to the alternative one presented at the end of this one, although I don’t know if it’s exactly the same. I’m quite sceptical about the correctness of my solution. I wanted to know whether this approach is correct or not.
My code:

Code
``````#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 = 5e4;
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);
}

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

int a[MX];
string b[60];
int last[60];

int main(){
NACHO();
int n, k; cin >> n >> k;
F0R(i, n) cin >> a[i];
F0R(i, k) cin >> b[i];
F0R(i, n){
F0R(j, k){
if(b[a[i]-1][j] == '1' && last[j] != -1) adj[i].pb(mp(last[j], i-last[j]));
}
last[a[i]-1] = i;
}
F0R(i, k) last[i] = -1;
R0F(i, n){
F0R(j, k){
if(b[a[i]-1][j] == '1' && last[j] != -1) adj[i].pb(mp(last[j], last[j]-i));
if(b[a[i]-1][j] == '1' && j == a[n-1]-1) adj[i].pb(mp(n-1, n-1-i));
}
last[a[i]-1] = i;
}
vb visited (n, 0);
vi dist (n, inf);
priority_queue<pi> q;
dist[0] = 0;
q.push(mp(0, 0));
while(sz(q)){
auto s = q.top().s;
q.pop();
if(visited[s]) continue;
visited[s] = 1;