计算几何 半平面交

LA 4992 && hdu 3761 Jungle Outpost

杭电的有点坑啊。。一直爆内存，后来发现大白的半平面交模板那里 point *p = new point[n]; line *q = new line[n]这里出了问题，应该是在函数里面申请不了比较大的数组，所以爆内存。。我在全局定义了两个数组就不会爆了。。

本来跑了17s多的，后来半平面交sort( l, l + n ) 被我注释了，就跑了9s多，LA上跑了 2s。。应该是输入数据比较好，不用按照极角排序。。然后就是照着大白的想法，二分答案，用半平面交判断是否满足条件。。

输入是按照逆时针输入的，所以用p[i] - p[(i+m+1)%n]表示直线的向量。。

这道题被坑了好久，好伤心。。贴的是在杭电ac的代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<vector>

using namespace std;

#define mnx 50050
#define LL long long
#define mod 1000000007
#define inf 0x3f3f3f3f
#define eps 1e-8
#define Pi acos(-1.0);
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1 | 1

int dcmp( double x ){
    if( fabs( x ) < eps ) return 0;
    return x < 0 ? -1 : 1;
}
struct point{
    double x, y;
    point( double x = 0, double y = 0 ) : x(x), y(y) {}
    point operator + ( const point &b ) const{
        return point( x + b.x, y + b.y );
    }
    point operator - ( const point &b ) const{
        return point( x - b.x, y - b.y );
    }
    point operator * ( const double &k ) const{
        return point( x * k, y * k );
    }
    point operator / ( const double &k ) const{
        return point( x / k, y / k );
    }
    bool operator < ( const point &b ) const{
        return dcmp( x - b.x ) < 0 || dcmp( x - b.x ) == 0 && dcmp( y - b.y ) < 0;
    }
    bool operator == ( const point &b ) const{
        return dcmp( x - b.x ) == 0 && dcmp( y - b.y ) == 0;
    }
    double len(){
        return sqrt( x * x + y * y );
    }
};
typedef point Vector;
struct line{
    point p; 
    Vector v;
    double ang;
    line() {}
    line( point p, point v ) : p(p), v(v) {
        ang = atan2( v.y, v.x );
    }
    bool operator < ( const line &b ) const{
        return ang < b.ang;
    }
};
double dot( Vector a, Vector b ){
    return a.x * b.x + a.y * b.y;
}
double cross( Vector a, Vector b ){
    return a.x * b.y - a.y * b.x;
}
bool onleft( line l, point p ){
    return cross( l.v, p - l.p ) > 0;
}
point getintersection( line a, line b ){
    Vector u = a.p - b.p;
    double t = cross( b.v, u ) / cross( a.v, b.v );
    return a.p + a.v * t;
}
point pp[mnx];
line q[mnx];
int halfplaneintersection( line *L, int n, point *poly ){
    //sort( L, L + n );
    int first, last;
    q[first = last = 0] = L[0];
    for( int i = 1; i < n; i++ ){
        while( first < last && !onleft( L[i], pp[last-1] ) ) last--;
        while( first < last && !onleft( L[i], pp[first] ) ) first++;
        q[++last] = L[i];
        if( fabs( cross( q[last].v, q[last-1].v ) ) < eps ){
            last--;
            if( onleft( q[last], L[i].p ) ) q[last] = L[i];
        }
        if( first < last ) pp[last-1] = getintersection( q[last-1], q[last] );
    }
    while( first < last && !onleft( q[first], pp[last-1] ) ) last--;
    if( last - first <= 1 ) return 0;
    pp[last] = getintersection( q[last], q[first] );
    int m = 0;
    for( int i = first; i <= last; i++ ){
        poly[m++] = pp[i];
    }
    return m;
}
point p[mnx], poly[mnx];
line l[mnx];
bool check( int n, int m ){
    if( n - m <= 2 ) return 1;
    for( int i = 0; i < n; i++ ){
        l[i] = line( p[i], p[i] - p[(i+m+1)%n] );
    }
    int all = halfplaneintersection( l, n, poly );
    if( !all ) return 1;
    else return 0;
}
int main(){
    int n, cas;
    scanf( "%d", &cas );
    while( cas-- ){
        scanf( "%d", &n );
        for( int i = 0; i < n; i++ ){
            scanf( "%lf %lf", &p[i].x, &p[i].y );
        }
        if( n > 50000 ) continue;
        int l = 0, r = n, ans;
        while( l < r ){
            int m = ( l + r ) >> 1;
            if( check( n, m ) == 1 ){
                ans = m, r = m;
            }
            else l = m + 1;
        }
        printf( "%d\n", ans );
    }
    return 0;
}