結果

問題 No.2376 障害物競プロ
ユーザー apricityapricity
提出日時 2023-07-08 12:15:16
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 504 ms / 4,000 ms
コード長 4,196 bytes
コンパイル時間 1,382 ms
コンパイル使用メモリ 133,856 KB
最終ジャッジ日時 2025-02-15 08:56:02
ジャッジサーバーID
(参考情報)
judge5 / judge2
このコードへのチャレンジ
(要ログイン)
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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<numeric>
#include<cmath>
#include<utility>
#include<tuple>
#include<cstdint>
#include<cstdio>
#include<iomanip>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<deque>
#include<unordered_map>
#include<unordered_set>
#include<bitset>
#include<cctype>
#include<chrono>
#include<random>
#include<cassert>
#include<cstddef>
#include<iterator>
#include<string_view>
#include<type_traits>
#ifdef LOCAL
# include "debug_print.hpp"
# define debug(...) debug_print::multi_print(#__VA_ARGS__, __VA_ARGS__)
#else
# define debug(...) (static_cast<void>(0))
#endif
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define FOR(i,a,b) for(int i=(a); i<(b); i++)
#define RFOR(i,a,b) for(int i=(b-1); i>=(a); i--)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define pb push_back
using ll = long long;
using D = double;
using LD = long double;
using P = pair<int, int>;
template<typename T> using PQ = priority_queue<T,vector<T>>;
template<typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
void yesno(bool flag) {cout << (flag?"Yes":"No") << "\n";}
template<typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
os << p.first << " " << p.second;
return os;
}
template<typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
is >> p.first >> p.second;
return is;
}
template<typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
int s = (int)v.size();
for (int i = 0; i < s; i++) os << (i ? " " : "") << v[i];
return os;
}
template<typename T>
istream &operator>>(istream &is, vector<T> &v) {
for (auto &x : v) is >> x;
return is;
}
void in() {}
template<typename T, class... U>
void in(T &t, U &...u) {
cin >> t;
in(u...);
}
void out() { cout << "\n"; }
template<typename T, class... U, char sep = ' '>
void out(const T &t, const U &...u) {
cout << t;
if (sizeof...(u)) cout << sep;
out(u...);
}
void outr() {}
template<typename T, class... U, char sep = ' '>
void outr(const T &t, const U &...u) {
cout << t;
outr(u...);
}
ll cross(ll x1, ll y1, ll x2, ll y2) {
return x1*y2-y1*x2;
}
ll dot(ll x1, ll y1, ll x2, ll y2) {
return x1*x2+y1*y2;
}
ll norm2(ll x, ll y){
return x*x + y*y;
}
int ccw(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3){
if(cross(x2-x1, y2-y1, x3-x1, y3-y1) > 0) return 1;
if(cross(x2-x1, y2-y1, x3-x1, y3-y1) < 0) return -1;
if(dot(x2-x1, y2-y1, x3-x1, y3-y1) < 0) return 2;
if(norm2(x2-x1, y2-y1) < norm2(x3-x1, y3-y1)) return -2;
return 0;
}
bool intersect(ll x1, ll y1, ll x2, ll y2, ll x3, ll y3, ll x4, ll y4){
int t1 = ccw(x1,y1,x2,y2,x3,y3) * ccw(x1,y1,x2,y2,x4,y4);
int t2 = ccw(x3,y3,x4,y4,x1,y1) * ccw(x3,y3,x4,y4,x2,y2);
return t1 < 0 and t2 < 0;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int n,m; in(n,m);
vector<ll> x1(n),x2(n),y1(n),y2(n);
rep(i,n) in(x1[i],y1[i],x2[i],y2[i]);
vector<ll> x, y;
rep(i,n){
x.pb(x1[i]);
x.pb(x2[i]);
y.pb(y1[i]);
y.pb(y2[i]);
}
const LD INF = 1e18;
vector<vector<LD>> dist(n*2, vector<LD> (n*2, INF));
rep(i,n*2) dist[i][i] = 0.0;
rep(i,2*n)FOR(j,i+1,2*n){
ll z1 = x[i], z2 = x[j], w1 = y[i], w2 = y[j];
int ok = 1;
rep(k,n){
ll p1 = x[2*k], p2 = x[2*k+1];
ll q1 = y[2*k], q2 = y[2*k+1];
if (intersect(z1,w1,z2,w2,p1,q1,p2,q2)) ok = 0;
}
if(ok) dist[i][j] = dist[j][i] = sqrtl(norm2(z1-z2,w1-w2));
}
rep(k,n*2)rep(i,n*2)rep(j,n*2){
chmin(dist[i][j], dist[i][k] + dist[k][j]);
}
while(m--){
int a,b,c,d; in(a,b,c,d); a--; b--; c--; d--;
int i = a*2+b, j = c*2+d;
printf("%.12Lf\n",dist[i][j]);
}
}
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