結果
問題 | No.2376 障害物競プロ |
ユーザー |
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提出日時 | 2023-07-25 12:47:42 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 899 ms / 4,000 ms |
コード長 | 3,589 bytes |
コンパイル時間 | 2,369 ms |
コンパイル使用メモリ | 204,292 KB |
最終ジャッジ日時 | 2025-02-15 18:56:36 |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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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> using namespace std; using ll = long long; // vector OA = (x, y) template <typename T> struct vec{ T x, y; vec (T xx=0, T yy=0) : x(xx), y(yy) {}; vec operator-() const { return vec(-x, -y); } vec& operator+=(const vec &w) { x += w.x; y += w.y; return *this; } vec& operator-=(const vec &w) { x -= w.x; y -= w.y; return *this; } vec operator+(const vec &w) const { vec res(*this); return res += w; } vec operator-(const vec &w) const { vec res(*this); return res-=w; } }; //Inner product of vectors v and w template <typename T> T dot(vec<T> v, vec<T> w){ return v.x * w.x + v.y * w.y; } //Outer product of vector v and w template <typename T> T outer(vec<T> v, vec<T> w){ return v.x * w.y - w.x * v.y; } //size of triangle ABC template <typename T> T heron(vec<T> &a, vec<T> &b, vec<T> &c){ return abs(outer(b-a, c-a)) / 2; } //Distance between point a and b template <typename T> T distance(vec<T> &a, vec<T> &b){ return sqrt(dot(a-b, a-b)); } //Convex Hull(Smallest Convex set containing all given vecs) //Grahum Scan (O(NlogN)) template <typename T> vector<vec<T>> convex_hull(vector<vec<T>> P){ sort(P.begin(), P.end(), [](vec<T> &p1, vec<T> &p2) { if (p1.x != p2.x) return p1.x < p2.x; return p1.y < p2.y; }); int N=P.size(), k=0, t; vector<vec<T>> res(N*2); for (int i=0; i<N; i++){ while(k > 1 && outer(res[k-1]-res[k-2], P[i]-res[k-1]) <= 0) k--; res[k] = P[i]; k++; } t = k; for (int i=N-2; i>=0; i--){ while(k > t && outer(res[k-1]-res[k-2], P[i]-res[k-1]) <= 0) k--; res[k] = P[i]; k++; } res.resize(k-1); return res; } //distance between line PQ and point R template <typename T> T dist_line_point(vec<T> &p, vec<T> &q, vec<T> &r){ // ax+by+c=0 T a = (q.y-p.y), b = (p.x-q.x), c = -p.x*q.y + p.y*q.x; cout << a << " " << b << " " << c << endl; return abs(a*r.x+b*r.y+c) / sqrt(a*a+b*b); }; //judge if line AB intersects line CD template <typename T> bool intersect(vec<T> &a, vec<T> &b, vec<T> &c, vec<T> &d){ T s1, t1, s2, t2; s1 = outer(b-a, c-a); t1 = outer(b-a, d-a); s2 = outer(d-c, a-c); t2 = outer(d-c, b-c); if (s1 * t1 < 0 && s2 * t2 < 0) return 1; return 0; } int main(){ using ld = long double; ll N, Q, a, b, c, d, x, y; cin >> N >> Q; vector<vec<ll>> p(N*2); vector<vec<ld>> q(N*2); vector<vector<ld>> dist(N*2, vector<ld>(N*2, 1e18)); for (int i=0; i<N*2; i++){ cin >> x >> y; p[i] = vec(x, y); q[i] = vec((ld)x, (ld)y); } for (int i=0; i<N*2; i++){ dist[i][i] = 0; for (int j=i+1; j<N*2; j++){ bool f=1; for (int k=0; k<N; k++){ if (intersect(p[i], p[j], p[2*k], p[2*k+1])) f=0; } if (f){ dist[i][j] = dist[j][i] = distance(q[i], q[j]); } } } /* for (int i=0; i<N*2; i++){ for (int j=0; j<N*2; j++) cout << dist[i][j] << " "; cout << endl; } */ for (int k=0; k<N*2; k++){ for (int i=0; i<N*2; i++){ for (int j=0; j<N*2; j++){ dist[i][j] = min(dist[i][j], dist[i][k]+dist[k][j]); } } } while(Q){ Q--; cin >> a >> b >> c >> d; a--; b--; c--; d--; cout << setprecision(18) << dist[a*2+b][c*2+d] << endl; } return 0; }