POJ2079 Triangle

题目链接

问题分析

假的题目,假的数据……

不可能有少于\(O(n^2)\)的做法的,少于\(O(n^2)\)的做法是不可能的。

然而由于假的数据,凸包上的点只有不到\(3000\)个,所以\(n^2\)就好了……
秘技是语言选C++而不是G++

参考程序

#include <cmath>
#include <algorithm>
#include <cstdio>
#define LL long long
using namespace std;

const LL Maxn = 50010;
struct point {
    LL x, y;
    point() {}
    point( LL _x, LL _y ) : x( _x ), y( _y ) {}
    inline point operator - ( const point Other ) const {
        return point( x - Other.x, y - Other.y );
    }
    inline LL operator * ( const point Other ) const {
        return x * Other.y - y * Other.x;
    }
    inline LL Dis() const { return x * x + y * y; }
};
LL N, M;
point A[ Maxn ], B[ Maxn ];
bool Cmp( const point X, const point Y ) {
    return ( X - A[ 1 ] ) * ( Y - A[ 1 ] ) > 0 ||
        ( ( X - A[ 1 ] ) * ( Y - A[ 1 ] ) == 0 && ( X - A[ 1 ] ).Dis() < ( Y - A[ 1 ] ).Dis() );
}

int main() {
    scanf( "%lld", &N );
    while( N != -1 ) {
        for( LL i = 1; i <= N; ++i ) scanf( "%lld%lld", &A[ i ].x, &A[ i ].y );
        for( LL i = 2; i <= N; ++i ) 
            if( A[ i ].y < A[ 1 ].y || ( A[ i ].y == A[ 1 ].y && A[ i ].x < A[ 1 ].x ) )
                swap( A[ i ], A[ 1 ] );
        sort( A + 2, A + N + 1, Cmp );
        M = 1; B[ 1 ] = A[ 1 ];
        for( LL i = 2; i <= N; ++i ) {
            for( ; M > 1 && ( A[ i ] - B[ M - 1 ] ) * ( B[ M ] - B[ M - 1 ] ) >= 0; --M );
            B[ ++M ] = A[ i ];
        }
        if( M <= 2 ) {
            printf( "0.00\n" );
            scanf( "%lld", &N );
            continue;
        }
        LL Ans = 0;
        for( LL i = 1, j, k, l; i < M; ++i ) {
            j = k = i + 1;
            l = ( k + 1 > M ) ? 1 : k + 1;
            for( ; ( B[ j ] - B[ i ] ) * ( B[ l ] - B[ k ] ) >= 0; k = l, l = ( k + 1 > M ) ? 1 : k + 1 );
            Ans = max( Ans, ( B[ j ] - B[ i ] ) * ( B[ k ] - B[ i ] ) );
            for( ; j <= M; ++j ) {
                for( ; ( B[ j ] - B[ i ] ) * ( B[ l ] - B[ k ] ) >= 0; k = l, l = ( k + 1 > M ) ? 1 : k + 1 );
                Ans = max( Ans, ( B[ j ] - B[ i ] ) * ( B[ k ] - B[ i ] ) );
            }
        }
        printf( "%.2lf\n", 0.5 * Ans );
        scanf( "%lld", &N );
    }
    return 0;
}
posted @ 2019-08-09 08:21 chy_2003 阅读(...) 评论(...) 编辑 收藏