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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#352060#7730. Convex Checker_SheepsheepWA 50ms11188kbC++177.4kb2024-03-12 20:09:502024-03-12 20:09:51

Judging History

你现在查看的是最新测评结果

  • [2024-07-04 19:27:17]
  • hack成功,自动添加数据
  • (/hack/727)
  • [2024-07-04 19:17:30]
  • hack成功,自动添加数据
  • (/hack/726)
  • [2024-03-12 20:09:51]
  • 评测
  • 测评结果:WA
  • 用时:50ms
  • 内存:11188kb
  • [2024-03-12 20:09:50]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std ;
#define ll long long
const int inf = 1e8 ;
const int N = 3e6+9 ;
typedef long double LD ;
const LD eps = 1e-7 ;
const LD box = 1e7 ;
int sgn( LD x )
{
    return x > eps ? 1 : ( x < -eps ? -1 : 0 ) ;
}
struct point
{
    LD x , y , polar_angle;
    point operator + (const point &a) const
    {
        return{ x+a.x , y+a.y } ;
    }
    point operator - (const point &a) const
    {
        return{ x-a.x , y-a.y } ;
    }
    point operator * (const LD &a) const
    {
        return{ x*a , y*a } ;
    }
    point operator / (const LD &a) const
    {
        return{ x/a , y/a } ;
    }
    friend bool operator == ( const point &a , const point &b )
    {
        return sgn(a.x-b.x) == 0 && sgn(a.y-b.y) == 0 ;
    }

    friend ostream & operator << ( ostream &out , point &a )
    {
        out << "(" << a.x << "," << a.y << ")" ;
        return out ;
    }
    friend istream & operator >> ( istream &in , point &a )
    {
        in >> a.x >> a.y ;
        return in ;
    }
    int idx , hull_rank ;
};
struct line
{
    point s , t ;
    friend bool operator == ( const line &a , const line &b )
    {
        return (a.s==b.s&&a.t==b.t)||(a.s==b.t&&a.t==b.s) ;
    }
};
LD sqr( LD x )
{
    return x*x ;
}
LD mo( point x )
{
    return sqrtl( sqr(x.x) + sqr(x.y) ) ;
}
LD dis( const point &a , const point &b )
{
    return sqrtl( sqr(a.x-b.x) + sqr(a.y-b.y) ) ;
}
LD dot( const point &a , const point &b )
{
    return a.x*b.x+a.y*b.y;
}
LD det( const point &a , const point &b )
{
    // + : b 在 a 的逆时针方向
    return a.x*b.y-b.x*a.y ;
}
bool point_on_segment( const point &a , const line &l )
{
    if( l.s == l.t )
    {
        return a == l.s ;
    }
    return sgn( det(l.s-a,a-l.t) ) == 0 && sgn( dot(l.s-a,l.t-a) ) <= 0 ;
}
bool two_side( const point &a , const point &b , const line &c )
{
    // 点是否在线两边
    return sgn( det(a-c.s,c.t-c.s) ) * sgn( det(b-c.s,c.t-c.s) ) < 0 ;
}
bool segment_inter_judge( const line &a , const line &b )
{
    bool ok = 0 ;
    ok |= point_on_segment( b.s , a ) ;
    ok |= point_on_segment( b.t , a ) ;
    ok |= point_on_segment( a.s , b ) ;
    ok |= point_on_segment( a.t , b ) ;
    ok |= ( two_side(a.s,a.t,b)&&two_side(b.s,b.t,a) ) ;
    return ok ;
}
bool ray_inter_judge( const line &a , const line &b )
{
    //前提都不是退化的射线
    return sgn( det( a.t-a.s , b.t-b.s ) ) == 0 ? 0 : 1 ;
}
LD point_to_line( const point &p , const line &l )
{
    return fabs( det(l.t-l.s,p-l.s) )/dis(l.s,l.t) ;
}
LD point_to_segment( const point &p , const line &l )
{
    if( l.s == l.t ) return dis( p , l.s ) ;
    if( sgn( dot(l.s-p,l.t-l.s) )*sgn( dot(l.t-p,l.t-l.s) ) <= 0 ) return point_to_line(p,l) ;
    else return min( dis(p,l.s) , dis(p,l.t) ) ;
}
LD arg( const line &a , const line &b )
{
    LD res = dot( a.t-a.s , b.t-b.s ) ;
    res /= mo( a.t-a.s ) ;
    res /= mo( b.t-b.s ) ;
    return acos( res ) ;
}
point line_intersect( const line &a , const line &b )
{
    LD u = det( a.t-a.s,b.s-a.s ) ;
    LD v = det( a.t-a.s , b.t-a.s ) ;
    return ( b.s*v - b.t*u )/(v-u) ;
}
bool turn_left( const point &a , const point &b , const point &c )
{
    // a 是中间点,看看ac是不是在ab的左边(逆时针方向)
    return sgn( det( b-a , c-a ) ) > 0 ;
}
bool turn_left( const line &a , const line &b , const line &c )
{
    // bc的交点是不是在向量a的左侧
    return turn_left( a.s , a.t , line_intersect(b,c) ) ;
}
int point_quadrant( const point &a )
{
    if( sgn( a.x ) >= 0 && sgn( a.y ) >= 0 ) return 1 ;
    else if( sgn(a.x) < 0 && sgn( a.y ) >= 0 ) return 2 ;
    else if( sgn(a.x) < 0 && sgn( a.y ) < 0 ) return 3 ;
    else return 4 ;

}
vector<point> convex_hull( vector<point> a )
{
    vector<point>ret ;
    int a_size = a.size() , ret_size = 0 ;
    if( a_size <= 2 ) return a ;
    sort( a.begin() , a.end() , []( point x , point y ){ return x.x == y.x ? x.y < y.y : x.x < y.x ; } )  ;
    for( int i = 0 ; i < a_size ; i ++ )
    {
        while( ret_size > 1 && !turn_left( ret[ret_size-2] , ret[ret_size-1] , a[i] ) )
        {
            ret.pop_back() ; ret_size -- ;
        }
        ret.push_back( a[i] ) ; ret_size ++ ;
    }
    int fix = ret_size ;
    for( int i = a_size-2 ; i >= 0 ; i -- )
    {
        while( ret_size > fix && !turn_left( ret[ret_size-2] , ret[ret_size-1] , a[i] ) )
        {
            ret.pop_back() ; ret_size -- ;
        }
        ret.push_back( a[i] ) ; ret_size ++ ;
    }
    ret.pop_back() ; ret_size -- ;

    return ret ;
}
int half( const point &a ) //在上下哪个平面
{
    return a.y > 0 || ( a.y == 0 && a.x > 0 ? 1 : 0 );
}
bool is_para( const line &a , const line &b ) //是不是共线
{
    return sgn( det(a.t-a.s,b.t-b.s) ) == 0 ;
}
bool cmp( const line &a , const line &b )
{
    // (0,0) is org
    int sign = half( a.t-a.s ) - half( b.t-b.s ) ; //是否在同一半平面
    int dir = sgn( det(a.t-a.s,b.t-b.s) ) ; //是否共线

    if( sign == 0 && dir == 0 ) return sgn( det(a.t-a.s , b.t-a.s) ) < 0 ; //距离原点近的
    else return sign ? sign > 0 : dir > 0 ;
    //如果异面,上半平面优先
    //如果同面,逆时针排序
}
vector<point> hpi( vector<line> A , LD DX , LD DY )
{
    int siz_a = A.size() ;
    vector<line>h ;
    for( int i = 0 ; i < siz_a ; i ++ ) h.push_back( A[i] ) ;

    h.push_back( {{DX,DY} , {0,DY}} ) ;
    h.push_back( {{0,DY} , {0,0}} ) ;
    h.push_back( {{0,0} , {DX,0}} ) ;
    h.push_back( {{DX,0} , {DX,DY}} ) ;
    sort( h.begin() , h.end() , cmp ) ;

    vector<line> q( h.size()+10 ) ;
    int l = 0 , r = -1 ;
    for( auto &i : h )
    {
        while( l<r && !turn_left(i,q[r-1],q[r]) ) --r ;
        while( l<r && !turn_left(i,q[l],q[l+1]) ) ++l ;
        if( l <= r && is_para(i,q[r]) ) continue ;
        q[++r] = i ;
    }
    while( r-l>1 && !turn_left(q[l],q[r-1],q[r]) ) --r ;
    while( r-l>1 && !turn_left(q[r],q[l],q[l+1]) ) ++l ;

    if( r-l < 2 ) return {} ;
    vector<point> ret(r-l+1) ;
    for( int i = l ; i <= r ; i ++ )
        ret[i-l] = line_intersect( q[i] , q[i==r?l:i+1] ) ;
    return ret ;
}
bool on_box( point a )
{
    if( sgn(a.x-box) == 0 || sgn(a.y-box) == 0 ) return 1 ;
    return 0;
}
bool is_open( vector<point>ret )
{
    if( ret.size() <= 2 )
    for( auto &u : ret ) if( on_box(u) ) return 1 ;
    return 0 ;
}
void solve()
{
    int n ; cin >> n ;
    vector<point>a(n) ;
    for( int i = 0 ; i < n ; i ++ )
    {
        cin >> a[i] ;
        if( a[i].x == a[0].x )
        {
            if( a[i].y < a[0].y ) swap( a[i] , a[0] ) ;
        }
        else if( a[i].x < a[0].x ) swap( a[i] , a[0] ) ;
    }
    for( int i = 1 ; i < n ; i ++ )
    {
        a[i].polar_angle = atan2( a[i].y-a[0].y , a[i].x-a[0].x ) ;
    }

    sort( a.begin() , a.end() , [&]( point x , point y ){ return x.polar_angle < y.polar_angle ;} );

    bool ok = 1 ;
    for( int i = 1 ; i < n ; i ++ )
    {
        if( !turn_left( a[i-1] , a[i] , a[(i+1)%n] ) ) ok = 0 ;
    }

    if( ok ) cout << "Yes\n" ;
    else cout << "No\n" ;

}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int tt = 1 ; //cin >> tt ;
    while( tt-- ) solve() ;
    return 0 ;
}
/*
5 4.321
-2 -1 3 -2
1 6 3 -2
1 6 -2 -1
-3 4 3 3
-2 1 5 4
*/

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3712kb

input:

3
0 0
1 0
0 1

output:

Yes

result:

ok answer is YES

Test #2:

score: 0
Accepted
time: 0ms
memory: 3768kb

input:

4
0 0
0 1
1 1
1 0

output:

Yes

result:

ok answer is YES

Test #3:

score: 0
Accepted
time: 0ms
memory: 3696kb

input:

4
0 0
0 3
1 2
1 1

output:

Yes

result:

ok answer is YES

Test #4:

score: 0
Accepted
time: 0ms
memory: 3956kb

input:

3
0 0
0 0
0 0

output:

No

result:

ok answer is NO

Test #5:

score: 0
Accepted
time: 0ms
memory: 3760kb

input:

5
1 0
4 1
0 1
2 0
3 2

output:

No

result:

ok answer is NO

Test #6:

score: 0
Accepted
time: 0ms
memory: 3816kb

input:

5
0 0
1000000000 0
1000000000 500000000
1000000000 1000000000
0 1000000000

output:

No

result:

ok answer is NO

Test #7:

score: 0
Accepted
time: 0ms
memory: 3832kb

input:

5
0 0
1000000000 0
1000000000 499999999
1000000000 1000000000
0 1000000000

output:

No

result:

ok answer is NO

Test #8:

score: 0
Accepted
time: 0ms
memory: 3696kb

input:

5
0 0
999999999 0
1000000000 50000000
999999999 1000000000
0 1000000000

output:

Yes

result:

ok answer is YES

Test #9:

score: -100
Wrong Answer
time: 50ms
memory: 11188kb

input:

128312
5578014 410408218
5585076 410404717
5588011 410403262
5588473 410403033
5589740 410402405
5593295 410400643
5593751 410400417
5597248 410398684
5598935 410397848
5600618 410397014
5605185 410394751
5610514 410392111
5614281 410390245
5617263 410388768
5621142 410386847
5630840 410382045
56310...

output:

No

result:

wrong answer expected YES, found NO