제출 #386236

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
386236		Return_0	Graph (BOI20_graph)	C++17	100 / 100	175 ms	22756 KiB

Graph

#pragma GCC optimize("Ofast,unroll-loops")
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma,tune=native")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef int ll;
typedef long double ld;
typedef pair <ll, ll> pll;

#ifdef SINA
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1> void __f(const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << std::endl; }
template <typename Arg1, typename... Args> void __f (const char* names, Arg1&& arg1, Args&&... args) {
	const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...); }
#define dbg2(x, j, n) cout<< #x << " : "; output((j), (n), x, 1); cout.flush();
#else
#define dbg(...) 0
#define dbg2(x, j, n) 0
#endif
#define SZ(x) ((ll)((x).size()))
#define File(s, t) freopen(s ".txt", "r", stdin); freopen(t ".txt", "w", stdout);
#define input(j, n, a) for (int _i = (j); _i < (n)+(j); _i++) cin>> a[_i];
#define output(j, n, a, t) for (int _i = (j); _i < (n)+(j); _i++) cout<< a[_i] << (((t) && _i != (n)+(j)-1)? ' ' : '\n');
#define kill(x) return cout<< (x) << endl, 0
#define cl const ll
#define fr first
#define sc second
#define lc (v << 1)
#define rc (lc | 1)
#define mid ((l + r) >> 1)
#define All(x) (x).begin(), (x).end()

cl inf = sizeof(ll) == 4 ? (1e9 + 10) : (3e18), mod = 1e9 + 7, MOD = 998244353;

template <class A,class B> ostream& operator << (ostream& out,const pair<A,B>&a){return out<<'('<<a.first<<", "<<a.second<<')';}
template <class A> ostream& operator << (ostream& out, const vector<A> &a) {
	out<< '['; for (int i = -1; ++i < int(a.size());) out<< a[i] << (i + 1 < int(a.size()) ? ", " : ""); return out<<']'; }
template <class T, typename _t = less <T> > using Tree = tree <T, null_type, _t, rb_tree_tag, tree_order_statistics_node_update>;

#define SIK { cout<< "NO\n"; exit(0); }

struct num {
	ll x, y;
	num () : x(0), y(0) {}
	num (cl &x, cl &y) : x(x), y(y) {}

	friend bool operator == (const num &a, const num &b) { return (a.x == b.x && a.y == b.y); }
	friend num operator + (const num &a, const num &b) { return num(a.x + b.x, a.y + b.y); }
	friend num operator - (const num &a, const num &b) { return num(a.x - b.x, a.y - b.y); }
	friend num operator - (cl &a, const num &b) { return num(-b.x, a - b.y); }
	friend double operator & (const num &a, const num &b) {
		num t = a - b;
		return t.y / double(-t.x);
	}
};

cl N = 1e5 + 7;

ll n, m;
double val [N], X, cur, ans;
bool mark [N], seen [N];
vector <pll> adj [N];
vector <ll> vec;
num a [N];

void dfs (ll v) {
	seen[v] = 1;
	vec.push_back(-(a[v].y * a[v].x));
	for (auto [u, w] : adj[v]) {
		if (X != inf) return;
		if (seen[u]) {
			num _ = w - a[v];
			if (_ == a[u]) continue;
			if (_.x == a[u].x) SIK;
			X = _ & a[u];
			return;
		} else {
			a[u] = w - a[v];

			dfs(u);
		}
	}
}

void dfs2 (ll v) {
	mark[v] = 1;
	cur += val[v];
	for (auto [u, w] : adj[v]) {
		if (mark[u]) {
			if (abs((val[v] + val[u]) - w) > 1e-7) SIK;
		} else {
			val[u] = w - val[v];
			dfs2(u);
		}
	}
}

int main ()
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cout<< fixed << setprecision(7);

	cin>> n >> m;
	for (ll i = 1, v, u, w; i <= m; i++) {
		cin>> v >> u >> w;
		adj[v].push_back({u, w});
		adj[u].push_back({v, w});
	}

	for (ll i = 1; i <= n; i++) if (!mark[i]) {
		X = inf; a[i].x = 1; cur = 0; vec.clear();
		dfs(i);
		if (X == inf) {
			sort(All(vec));
			val[i] = vec[SZ(vec) / 2];
			dfs2(i);
			ans += cur;
		}
		else {
			val[i] = X;
			dfs2(i);
			ans += cur;
		}
	}

	cout<< "YES\n";
	output(1, n, val, 1);

	cerr<< "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";

	return 0;
}
/*
4 4
1 2 1
2 3 2
1 3 2
3 4 1

2 1
1 2 1

3 2
1 2 2
2 3 2

3 4
1 2 2
2 2 1
2 1 1
1 2 2

*/

• Subtask 1: 5 / 5
• Subtask 2: 12 / 12
• Subtask 3: 17 / 17
• Subtask 4: 24 / 24
• Subtask 5: 42 / 42