結果

問題 No.1069 電柱 / Pole (Hard)
ユーザー heno239
提出日時 2020-08-02 19:53:54
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 757 ms / 2,000 ms
コード長 5,833 bytes
コンパイル時間 2,143 ms
コンパイル使用メモリ 151,396 KB
実行使用メモリ 62,348 KB
最終ジャッジ日時 2024-07-23 15:09:07
合計ジャッジ時間 11,773 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 79
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 998244353;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);

ll mod_pow(ll x, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * x % m;
		x = x * x % m; n >>= 1;
	}
	return res;
}
struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n % mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a * a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}

struct edge {
	int to; ld cost;
};
using speP = pair<ld, int>;
ld memo[2000][2000];
void solve() {
	int n, m,k; cin >> n >> m>>k;
	vector<vector<edge>> G(n);
	vector<int> x(n), y(n);
	int s, t; cin >> s >> t; s--; t--;
	rep(i, n)cin >> x[i] >> y[i];
	rep(i, n)Rep(j, i + 1, n) {
		ld dist = sqrtl(pow(x[j] - x[i], 2) + pow(y[j] - y[i],2));
		memo[i][j] = memo[j][i] = dist;
	}
	rep(i, m) {
		int p, q; cin >> p >> q; p--; q--;
		ld dist = sqrtl(pow(x[p] - x[q], 2) + pow(y[p] - y[q], 2));
		G[p].push_back({ q,dist });
		G[q].push_back({ p,dist });
	}
	vector<ld> dist(n, INF);
	vector<int> par(n);
	dist[s] = 0;
	priority_queue<speP, vector<speP>, greater<speP>> q;
	/*q.push({ 0,s });
	while (!q.empty()) {
		speP p = q.top(); q.pop();
		int id = p.second;
		if (dist[id] < p.first)continue;
		for (edge e : G[id]) {
			ld nd = e.cost + dist[id];
			if (nd < dist[e.to]) {
				par[e.to] = id;
				dist[e.to] = nd;
				q.push({ nd,e.to });
			}
		}
	}*/
	map<vector<int>, bool> chked;
	auto calc = [&](vector<int> v)->pair<ld, vector<int>> {
		if (chked[v])return { -1,{} };
		chked[v] = true;
		vector<bool>ban(n, false);
		for (int id : v)ban[id] = true;
		fill(all(dist), INF); dist[v.back()] = 0;
		q.push({ 0,v.back() });
		while (!q.empty()) {
			speP p = q.top(); q.pop();
			int id = p.second;
			if (dist[id] < p.first)continue;
			for (edge e : G[id]) {
				if (ban[e.to])continue;
				ld nd = e.cost + p.first;
				if (nd < dist[e.to]) {
					dist[e.to] = nd;
					par[e.to] = id;
					q.push({ nd,e.to });
				}
			}
		}
		if (dist[t] == INF)return { -1,{} };
		vector<int> res2 = v;
		vector<int> las;
		int cur = t;
		while (cur != v.back()) {
			las.push_back(cur);
			cur = par[cur];
		}
		per(i, (int)las.size())res2.push_back(las[i]);
		return { dist[t],res2 };
	};

	pair<ld, vector<int>> cur = calc({ s });
	vector<ld> ans(k, -1);
	ans[0] = cur.first;

	vector<vector<int>> pre;
	pre.push_back(cur.second);

	set<pair<ld, vector<int>>> st;
	rep1(aa, k-1) {
		vector<int> las = pre.back();
		//cout << "??? " << las.size() << "\n";
		vector<int> ids(pre.size());
		rep(i, pre.size())ids[i] = i;

		vector<int> vc;
		ld s = 0;

		vector<bool> ban(n, false);
		rep(i, (int)las.size() - 1) {
			ban[las[i]] = true;
			vector<int> nex;
			for (int id : ids) {
				if (pre[id][i] == las[i])nex.push_back(id);
			}
			swap(nex, ids);
			//cout << "! " << ids.size() << "\n";
			vc.push_back(las[i]);
			for (edge e : G[las[i]]) {
				bool valid = true;
				if (ban[e.to])valid = false;
				for (int id : ids)if (e.to == pre[id][i + 1])valid = false;
				if (valid) {
					//cout << "? " << i << " " << e.to << "\n";
					vc.push_back(e.to);

					pair<ld, vector<int>> p = calc(vc);
					if (p.first >= 0) {
						p.first += s+memo[las[i]][e.to];
						st.insert(p);
					}

					vc.pop_back();
				}
			}
			s += memo[las[i]][las[i + 1]];
		}
		if (st.empty())break;
		pair<ld, vector<int>> ne = *st.begin(); st.erase(st.begin());
		ans[aa] = ne.first;
		pre.push_back(ne.second);
	}
	rep(i, k) {
		if (ans[i] < 0)cout << -1 << "\n";
		else cout << ans[i] << "\n";
	}
}




signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed << setprecision(10);
	//init_f();
	//cout << grandy(2, 3, false, false) << "\n";
	//int t; cin >> t; rep(i, t)
		solve();
	return 0;
}
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