Submission #879786

#	Time	Username	Problem	Language	Result	Execution time	Memory
879786		phoenix0423	timeismoney (balkan11_timeismoney)	C++17	100 / 100	448 ms	1108 KiB

timeismoney

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pll;
#define fastio ios::sync_with_stdio(false), cin.tie(0)
#pragma GCC optimize("Ofast")
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define REP1(i, n) FOR(i, 0, n + 1)
#define pb push_back
#define pf push_front
#define eb emplace_back
#define f first
#define s second
#define lowbit(x) x&-x
#define ckmin(a, b) a = min(a, b)
#define ckmax(a, b) a = max(a, b)
#define int long long 
const int INF = 1e9 + 7;
const int maxn = 300;
const int maxm = 1e4 + 5;
int n, m;
int par[maxn];
int root(int x){ return x == par[x] ? x : par[x] = root(par[x]);}

struct edge{
	edge(){}
	edge(int _x, int _y, int _t, int _c, int _w) : x(_x), y(_y), t(_t), c(_c), w(_w){}
	int x, y;
	int t, c;
	int w;
	bool operator < (const edge& other) const{
		return w < other.w;
	}
} e[maxm];

struct pt{
	int x, y; // (T, C)
	pt(){}
	pt(int _x, int _y) : x(_x), y(_y){}
	pt operator - (const pt b){
		return pt(x - b.x, y - b.y);
	}
	int cross(const pt b){
		return x * b.y - y * b.x;
	}
} ans, A, B;
vector<pll> best;

pt kruskal(){
	for(int i = 0; i < n; i++) par[i] = i;
	sort(e, e + m);
	pt ret = {0, 0};
	vector<pll> cur;
	for(int i = 0; i < m; i++){
		int x = root(e[i].x), y = root(e[i].y);
		if(x == y) continue;
		cur.eb(e[i].x, e[i].y);
		par[x] = y, ret.x += e[i].t, ret.y += e[i].c;
	}
	if(ret.x * ret.y < ans.x * ans.y) ans = ret, best = cur;
	return ret;
}
void work(pt A, pt B){ // divide and conquer
	for(int i = 0; i < m; i++) e[i].w = e[i].t * (A.y - B.y) + e[i].c * (B.x - A.x);
	pt C = kruskal();
	if((B - A).cross(C - A) >= 0) return;
	work(A, C), work(C, B); 
}
signed main(void){
	fastio;
	cin>>n>>m;
	ans = {INF, 1};
	for(int i = 0; i < m; i++){
		int x, y, t, c;
		cin>>x>>y>>t>>c;
		e[i] = edge(x, y, t, c, 0);
	}
	for(int i = 0; i < m; i++) e[i].w = e[i].t;
	A = kruskal();
	for(int i = 0; i < m; i++) e[i].w = e[i].c;
	B = kruskal();
	work(A, B);
	cout<<ans.x<<" "<<ans.y<<"\n";
	for(auto [x, y] : best) cout<<x<<" "<<y<<"\n";
}

• Subtask 1: 100 / 100