結果
問題 | No.2955 Pizza Delivery Plan |
ユーザー |
|
提出日時 | 2025-02-07 14:31:10 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 6,016 bytes |
コンパイル時間 | 3,791 ms |
コンパイル使用メモリ | 315,876 KB |
実行使用メモリ | 40,104 KB |
最終ジャッジ日時 | 2025-02-07 14:32:17 |
合計ジャッジ時間 | 62,771 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 9 TLE * 19 |
ソースコード
#include <bits/stdc++.h>using namespace std;#define rep(i, n) for(int i=0; i<n; i++)#define debug 0#define YES cout << "Yes" << endl;#define NO cout << "No" << endl;using ll = long long;using ld = long double;const int mod = 998244353;const int MOD = 1000000007;const double pi = atan2(0, -1);const int inf = 1 << 31 - 1;const ll INF = 1LL << 63 - 1;#include <time.h>#include <chrono>//vectorの中身を空白区切りで出力template<typename T>void printv(vector<T> v) {for (int i = 0; i < v.size(); i++) {cout << v[i];if (i < v.size() - 1) {cout << " ";}}cout << endl;}//vectorの中身を改行区切りで出力template<typename T>void print1(vector<T> v) {for (auto x : v) {cout << x << endl;}}//二次元配列を出力template<typename T>void printvv(vector<vector<T>> vv) {for (vector<T> v : vv) {printv(v);}}//vectorを降順にソートtemplate<typename T>void rsort(vector<T>& v) {sort(v.begin(), v.end());reverse(v.begin(), v.end());}//昇順priority_queueを召喚template<typename T>struct rpriority_queue {priority_queue<T, vector<T>, greater<T>> pq;void push(T x) {pq.push(x);}void pop() {pq.pop();}T top() {return pq.top();}size_t size() {return pq.size();}bool empty() {return pq.empty();}};//mod mod下で逆元を算出する//高速a^n計算(mod ver.)ll power(ll a, ll n) {if (n == 0) {return 1;}else if (n % 2 == 0) {ll x = power(a, n / 2);x *= x;x %= mod;return x;}else {ll x = power(a, n - 1);x *= a;x %= mod;return x;}}//フェルマーの小定理を利用ll modinv(ll p) {return power(p, mod - 2) % mod;}//Mexを求めるstruct Mex {map<int, int> mp;set<int> s;Mex(int Max) {for (int i = 0; i <= Max; i++) {s.insert(i);}}int _mex = 0;void Input(int x) {mp[x]++;s.erase(x);if (_mex == x) {_mex = *begin(s);}}void Remove(int x) {if (mp[x] == 0) {cout << "Mex ERROR!: NO VALUE WILL BE REMOVED" << endl;}mp[x]--;if (mp[x] == 0) {s.insert(x);if (*begin(s) == x) {_mex = x;}}}int mex() {return _mex;}};//条件分岐でYes/Noを出力するタイプのやつvoid YN(bool true_or_false) {cout << (true_or_false ? "Yes" : "No") << endl;}//Union-Findstruct UnionFind {vector<int> par;UnionFind(int N) : par(N) {for (int i = 0; i < N; i++) {par[i] = -1;}}//root(x):xの根を求める関数int root(int x) {if (par[x] == -1) {return x;}else {return par[x] = root(par[x]);}}//isSame(x,y):xとyが同じグループならtrueを返す関数bool isSame(int x, int y) {if (root(x) == root(y)) {return true;}else {return false;}}//Union(x,y):xとyの根をつなげる関数void Union(int x, int y) {int X = root(x);int Y = root(y);if (X == Y) {return;}if (X < Y) {swap(X, Y);}par[X] = Y;}};//最大公約数(ユークリッドの互除法)ll gcd(ll a, ll b) {if (b > a) {swap(a, b);}while (a % b != 0) {ll t = a;a = b;b = t % b;}return b;}//最小公倍数(gcdを定義しておく)ll lcm(ll a, ll b) {ll g = gcd(a, b);ll x = (a / g) * b;return x;}struct SegTree {vector<int> tree;vector<int> lazy;int n = 1;SegTree(int N, int H) {while (n < N) {n *= 2;}vector<int> v(n * 2, H + 1);swap(tree, v);vector<int> u(n * 2, 200005);swap(lazy, u);}void update(int val, int l, int r, int u, int t, int now) {if (l >= t || r <= u) {return;}else if (l <= u && r >= t) {lazy[now] = min(lazy[now], val);}else {tree[now] = min(tree[now], val);update(val, l, r, u, (u + t) / 2, now * 2 + 1);update(val, l, r, (u + t) / 2, t, now * 2 + 2);}}void treeup(int now, int val) {if (tree[now] > val) {tree[now] = val;if (now > 0) {treeup((now - 1) / 2, val);}}}int Min(int l, int r, int u, int t, int now) {if (l >= t || r <= u) {return inf;}else if (l <= u && r >= t) {treeup(now, lazy[now]);return tree[now];}else {treeup(now, lazy[now]);lazy[now * 2 + 1] = min(lazy[now * 2 + 1], lazy[now]);lazy[now * 2 + 2] = min(lazy[now * 2 + 2], lazy[now]);return min(Min(l, r, u, (u + t) / 2, now * 2 + 1), Min(l, r, (u + t) / 2, t, now * 2 + 2));}}int query(int l, int r) {return Min(l, r, 0, n, 0);}};int N, K, m=0;vector<vector<int>> group;vector<vector<ld>> plot;ld ans = INF;ld dist(vector<ld> x, vector<ld> y) {ld d = sqrt((x[0] - y[0]) * (x[0] - y[0]) + (x[1] - y[1]) * (x[1] - y[1]));return d;}void calc() {ld score = 0;for (vector<int> g : group) {map<ll, map<int, ld>> record;rpriority_queue < tuple<ld, ll, int>> pq;record[0][0] = 0;pq.push({ 0,0,0 });while (!pq.empty()) {tuple<ld,ll,int> tmp = pq.top();pq.pop();ld now_score;bitset<15> b;int now;tie(now_score, b, now) = tmp;for (int x : g) {if (!b[x] && now != x) {b[x] = 1;ld d = dist(plot[now], plot[x]);if (!record[b.to_ullong()].count(x) || record[b.to_ullong()][x] > now_score + d) {record[b.to_ullong()][x] = now_score + d;pq.push({ now_score + d, b.to_ullong(), x});}b[x] = 0;}}}bitset<15> e = 0;for (int x : g) {e[x] = 1;}score += record[e.to_ullong()][0];}ans = min(ans, score);}void dfs(int i) {if (i <= N) {for (int j = 0; j < group.size(); j++) {if (group[j].size() <= K) {group[j].push_back(i);dfs(i + 1);group[j].pop_back();}}if (group.size() < m) {group.push_back({ 0, i });dfs(i + 1);group.pop_back();}}else {calc();}}int main() {cin >> N >> K;for (int i = 0; i < N; i += K) {m++;}plot.push_back({ 0,0 });rep(i, N) {ld x, y;cin >> x >> y;plot.push_back({ x,y });}dfs(1);cout << setprecision(10) << fixed;cout << ans << endl;}