# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
305551 | AmineWeslati | 자매 도시 (APIO20_swap) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//Never stop trying
#include "swap.h"
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define boost ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef string str;
typedef long long ll;
#define int ll
typedef double db;
typedef long double ld;
typedef pair<int, int> pi;
#define fi first
#define se second
typedef vector<int> vi;
typedef vector<pi> vpi;
typedef vector<str> vs;
typedef vector<ld> vd;
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define endl "\n"
#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)
const int MOD = 1e9 + 7; //998244353
const ll INF = 1e18;
const int MX = 1e5 + 10;
const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //right left down up
template<class T> using V = vector<T>;
template<class T> bool ckmin(T& a, const T& b) { return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); } // divide a by b rounded up
constexpr int log2(int x) { return 31 - __builtin_clz(x); } // floor(log2(x))
#define dbg(x) cerr << " - " << #x << " : " << x << endl;
#define dbgs(x,y) cerr << " - " << #x << " : " << x << " / " << #y << " : " << y << endl;
#define dbgv(v) cerr << " - " << #v << " : " << endl << "[ "; for(auto it : v) cerr << it << ' '; cerr << ']' << endl;
void IO() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
int N,M;
vpi adj[MX];
void init(int NN, int MM, vi U, vi V, vi W){
N=NN,M=MM;
FOR(i,0,M){
adj[U[i]].pb({V[i],W[i]});
adj[V[i]].pb({U[i],W[i]});
}
}
int getMinimumFuelCapacity(int X, int Y) {
if(N-1==M) return -1;
return 0;
}
/* tree
5 4
1 2 1
1 3 1
2 4 1
2 5 1
*/
/* Careful!!!
.Array bounds
.Infinite loops
.Uninitialized variables / empty containers
.Order of input
Some insights:
.Binary search
.Graph representation
.Write brute force code
.Change your approach
*/