結果

問題 No.245 貫け!
ユーザー torus711
提出日時 2015-07-17 23:38:13
言語 C++11
(gcc 13.3.0)
結果
WA  
実行時間 -
コード長 4,555 bytes
コンパイル時間 770 ms
コンパイル使用メモリ 97,292 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-08 09:39:18
合計ジャッジ時間 1,693 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
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 * 12 WA * 4
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 <iomanip>
#include <sstream>
#include <vector>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iterator>
#include <limits>
#include <numeric>
#include <utility>
#include <cmath>
#include <cassert>
#include <cstdio>
using namespace std; using namespace placeholders;
using LL = long long;
using ULL = unsigned long long;
using VI = vector< int >;
using VVI = vector< vector< int > >;
using VS = vector< string >;
using SS = stringstream;
using PII = pair< int, int >;
using VPII = vector< pair< int, int > >;
template < typename T = int > using VT = vector< T >;
template < typename T = int > using VVT = vector< vector< T > >;
template < typename T = int > using LIM = numeric_limits< T >;
template < typename T > inline istream& operator>>( istream &s, vector< T > &v ){ for ( T &t : v ) { s >> t; } return s; }
template < typename T > inline ostream& operator<<( ostream &s, const vector< T > &v ){ for ( int i = 0; i < int( v.size() ); ++i ){ s << ( " " + !i
    ) << v[i]; } return s; }
template < typename T > inline T fromString( const string &s ) { T res; istringstream iss( s ); iss >> res; return res; };
template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); };
#define REP2( i, n ) REP3( i, 0, n )
#define REP3( i, m, n ) for ( int i = ( int )( m ); i < ( int )( n ); ++i )
#define GET_REP( a, b, c, F, ... ) F
#define REP( ... ) GET_REP( __VA_ARGS__, REP3, REP2 )( __VA_ARGS__ )
#define FOR( e, c ) for ( auto &e : c )
#define ALL( c ) ( c ).begin(), ( c ).end()
#define AALL( a, t ) ( t* )a, ( t* )a + sizeof( a ) / sizeof( t )
#define DRANGE( c, p ) ( c ).begin(), ( c ).begin() + ( p ), ( c ).end()
#define SZ( v ) ( (int)( v ).size() )
#define PB push_back
#define EM emplace
#define EB emplace_back
#define BI back_inserter
#define EXIST( c, e ) ( ( c ).find( e ) != ( c ).end() )
#define MP make_pair
#define fst first
#define snd second
#define DUMP( x ) cerr << #x << " = " << ( x ) << endl
#include <complex>
using Point = complex< double >;
constexpr double EPS = 1e-8;
constexpr double PI = acos( -1 );
constexpr Point O( 0, 0 );
// Point
istream& operator>> ( istream &s, Point &a )
{
double r, i;
s >> r >> i;
a = Point( r, i );
return s;
}
//
double dot( const Point &a, const Point &b )
{
return a.real() * b.real() + a.imag() * b.imag();
}
//
double cross( const Point &a, const Point &b )
{
return a.real() * b.imag() - a.imag() * b.real();
}
// p ( q1, q2 )
// include : dot, cross
bool intersectPS( const Point &p, const Point &q1, const Point &q2 )
{
// p 0 ⇔ ( q1, q2 ) p
// p 0 p q1, q2
return abs( cross( q1 - p, q2 - p ) ) <= EPS && dot( q1 - p, q2 - p ) <= EPS;
}
// ( p1, p2 ) ( q1, q2 )
// include : cross
Point intersectionLL( const Point &p1, const Point &p2, const Point &q1, const Point &q2 )
{
return p1 + ( p2 - p1 ) * ( cross( q2 - q1, q1 - p1 ) / cross( q2 - q1, p2 - p1 ) );
}
// ( p1, p2 ) ( q1, q2 )
// include : cross, intersectPS, intersectionLL
bool intersectSS( const Point &p1, const Point &p2, const Point &q1, const Point &q2 )
{
if ( abs( cross( p1 - p2, q1 - q2 ) ) <= EPS ) //
{
//
return intersectPS( q1, p1, p2 ) ||
intersectPS( q2, p1, p2 ) ||
intersectPS( p1, q1, q2 ) ||
intersectPS( p2, q1, q2 );
}
else
{
//
Point r( intersectionLL( p1, p2, q1, q2 ) );
return intersectPS( r, p1, p2 ) && intersectPS( r, q1, q2 );
}
}
int main()
{
cin.tie( 0 );
ios::sync_with_stdio( false );
int N;
cin >> N;
VT< Point > ps( N * 2 );
cin >> ps;
int res = 0;
REP( i, N * 2 )
{
REP( j, i )
{
const Point a = ( ps[j] - ps[i] ) * 300.;
const Point p = ps[i] - a, q = ps[i] + a;
int tmp = 0;
for ( int k = 0; k < 2 * N; k += 2 )
{
tmp += intersectSS( p, q, ps[k], ps[ k + 1 ] );
}
res = max( res, tmp );
}
}
cout << res << endl;
return 0;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0