結果

問題 No.2376 障害物競プロ
ユーザー srjywrdnprktsrjywrdnprkt
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
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 * 40
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 #
プレゼンテーションモードにする

#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;
}
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