POJ 2653 线段相交判断 （叉积）

题目链接：http://poj.org/problem?id=2653

题解：线段相交判断 叉积基本运用

#include<cstdio>
#include<cstring>
#include<cmath>
#define N 110000
bool ans
;
//定义点
struct Point
{
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y) {}
};

struct segment
{
Point a, b;
}seg
;

typedef Point Vector; //Vector 为 Point的别名

//向量+向量=向量 点+向量=点
Vector operator + (Vector A, Vector B) {return Vector(A.x+B.x, A.y+B.y);}

//点-点=向量
Vector operator - (Point A, Point B) {return Vector(A.x-B.x, A.y-B.y);}

//向量*数=向量
Vector operator * (Vector A, double p) {return Vector(A.x*p, A.y*p);}

//向量/数=向量
Vector operator / (Vector A, double p) {return Vector(A.x/p, A.y/p);}

bool operator < (const Point & a, const Point & b)
{
return a.x < b.x || (a.x == b.x && a.y < b.y);
}

const double eps = 1e-10;
int dcmp(double x)
{
if(fabs(x) < eps) return 0;
else return x < 0 ? -1 : 1;
}

//点积：两者长度乘积再乘上夹角余弦 XaXb + YaYb
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;
}

//线段相交判定（原理：大白书258）

//每条线段的两个端点都在另一条线段的两侧
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)//a,b线段
{
double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1),
c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);

return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4) < 0;
}

int main ()
{
int n;
while(scanf("%d", &n), n)
{
memset(ans, 0, sizeof(ans));

for(int i = 1; i <= n ;i++)
scanf("%lf %lf %lf %lf", &seg[i].a.x, &seg[i].a.y, &seg[i].b.x, &seg[i].b.y);

for(int i = 1; i <= n; i++)
for(int j = i+1; j <= n; j++)
{
if(SegmentProperIntersection(seg[i].a, seg[i].b, seg[j].a, seg[j].b))
{
ans[i] = 1;
break;
}
}

int flag = 0;
printf("Top sticks:");
for(int i = 1; i <= n; i++)
if(ans[i] == 0)
{
if(flag == 0)
flag = 1;
else
printf(",");
printf(" %d", i);

}
printf(".\n");

}

return 0;
}