結果
問題 | No.2376 障害物競プロ |
ユーザー |
|
提出日時 | 2023-07-07 23:03:04 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 835 ms / 4,000 ms |
コード長 | 2,241 bytes |
コンパイル時間 | 3,819 ms |
コンパイル使用メモリ | 256,044 KB |
最終ジャッジ日時 | 2025-02-15 08:17:49 |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 40 |
ソースコード
#include<bits/stdc++.h>#include<atcoder/all>using namespace std;using namespace atcoder;using ll = long long;using vll = vector<ll>;using vvll = vector<vll>;using mint = modint998244353;using vb = vector<bool>;using vvb = vector<vb>;using vvvb = vector<vvb>;using vd = vector<double>;using vvd = vector<vd>;using vvvd = vector<vvd>;#define rep(i,n) for(ll i=(ll)(0); i<(ll(n)); ++i)#define all(x) (x).begin(), (x).end()vector<double> X, Y;double dis(ll i, ll j) {double dx = X[i] - X[j];double dy = Y[i] - Y[j];return sqrt(dx * dx + dy * dy);}const double EPS = 1e-12;bool cross(ll a, ll b, ll c, ll d) {double s, t;s = (X[a] - X[b]) * (Y[c] - Y[a]) - (Y[a] - Y[b]) * (X[c] - X[a]);t = (X[a] - X[b]) * (Y[d] - Y[a]) - (Y[a] - Y[b]) * (X[d] - X[a]);if (s * t > -EPS)return false;s = (X[c] - X[d]) * (Y[a] - Y[c]) - (Y[c] - Y[d]) * (X[a] - X[c]);t = (X[c] - X[d]) * (Y[b] - Y[c]) - (Y[c] - Y[d]) * (X[b] - X[c]);if (s * t > -EPS)return false;return true;}void warshall_floyd(int n, vvd& d) {for (int k = 0; k < n; k++) { // 経由する頂点for (int i = 0; i < n; i++) { // 始点for (int j = 0; j < n; j++) { // 終点d[i][j] = min(d[i][j], d[i][k] + d[k][j]);}}}}int main() {ll N, M;cin >> N >> M;vvd D(2 * N, vd(2 * N, 1e18));rep(i, 2 * N)D[i][i] = 0;X.resize(2 * N);Y.resize(2 * N);rep(i, 2 * N)cin >> X[i] >> Y[i];rep(i, N) {bool OK = 1;//re//D[2 * i][2 * i + 1] = D[2 * i + 1][2 * i] = dis(2 * i, 2 * i + 1);}rep(j, 2 * N) {rep(k, 2 * N) {bool OK = 1;rep(i, N) {if (j == k || j / 2 == i || k / 2 == i)continue;if (cross(2 * i, 2 * i + 1, j, k))OK = 0;}if (OK)D[j][k] = D[k][j] = dis(k, j);}}warshall_floyd(2 * N, D);cout << fixed << setprecision(16);rep(i, M) {ll A, B, C, DD;cin >> A >> B >> C >> DD;A--; C--;A = 2 * A + B - 1;C = 2 * C + DD - 1;cout << D[A][C] << endl;}}