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HDU 1086 - You can Solve a Geometry Problem too

2016-07-27 22:30 267 查看
判断线段两两相交的个数

打模板熟练程度++;

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double EPS = 1e-10;
struct Point
{
double x,y;
Point(){}
Point(double x1,double y1):x(x1),y(y1) {}
}a[105],b[105];
typedef Point Vect;
Vect operator - (Vect A, Vect B)
{
return Vect(A.x - B.x, A.y - B.y);
}
int dcmp(double x)
{
return fabs(x) < EPS ? 0 : (x < 0? -1: 1);
}
double Cross(Vect A, Vect B)
{
return A.x * B.y - A.y * B.x;
}
double Dot(Vect A, Vect B)
{
return A.x * B.x + A.y * B.y;
}
bool OnSegment(Point p, Point a1, Point a2)
{
return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp( Dot(a1 - p, a2 - p) ) <= 0;
}

bool SegmentIntersection(Point a1, Point a2, Point b1, Point b2)
{
double c1 = Cross(a2 - a1, b1 - a1);
double c2 = Cross(a2 - a1, b2 - a1);
double c3 = Cross(b2 - b1, a1 - b1);
double c4 = Cross(b2 - b1, a2 - b1);
if(dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0) return 1;
else if(OnSegment(b1, a1, a2) ) return 1;
else if(OnSegment(b2, a1, a2) ) return 1;
else if(OnSegment(a1, b1, b2) ) return 1;
else if(OnSegment(a2, b1, b2) ) return 1;
else return 0;
}

int main()
{
int n;
while(~scanf("%d",&n) && n)
{
for(int i = 1; i <= n; i++)
scanf("%lf%lf%lf%lf", &a[i].x, &a[i].y, &b[i].x, &b[i].y);
int ans = 0;
for(int i = 1; i <= n; i++)
for(int j = i + 1; j <= n; j++)
if(SegmentIntersection(a[i], b[i], a[j], b[j]))
ans++;
printf("%d\n", ans);
}
}
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