hdu 1086 You can Solve a Geometry Problem too 求n条直线交点的个数
You can Solve a Geometry Problem too
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13549 Accepted Submission(s): 6645
Give you N (1<=N<=100) segments(线段), please output the number of all intersections(交点). You should count repeatedly if M (M>2) segments intersect at the same point.
Note:
You can assume that two segments would not intersect at more than one point.
A test case starting with 0 terminates the input and this test case is not to be processed.
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<vector>
#include<utility>
#include<map>
#include<queue>
#include<set>
#define mx 0x3f3f3f3f
#define ll long long
using namespace std;
const int N = ;
int flag;
double ans1,ans2,yy;
struct Point//定义点的结构体
{
double x, y;
};
struct stline//定义边的结构体
{
Point a, b;
} line[]; bool cmp(Point a, Point b)
{
return a.y < b.y;
}
int dblcmp(double a, double b)
{
if (fabs(a - b) <= 1E-) return ;
if (a > b) return ;
else return -;
}
//***************点积判点是否在线段上***************
double dot(double x1, double y1, double x2, double y2) //点积
{
return x1 * x2 + y1 * y2;
} int point_on_line(Point a, Point b, Point c) //求a点是不是在线段bc上,>0不在,=0与端点重合,<0在。
{
return dblcmp(dot(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y), );
}
//**************************************************
double cross(double x1, double y1, double x2, double y2)
{
return x1 * y2 - x2 * y1;
}
double ab_cross_ac(Point a, Point b, Point c) //ab与ac的叉积
{
return cross(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y);
}
int ab_cross_cd (Point a,Point b,Point c,Point d) //求ab是否与cd相交,交点为p。1规范相交,0交点是一线段的端点,-1不相交。
{
double s1,s2,s3,s4;
int d1,d2,d3,d4;
Point p;
d1=dblcmp(s1=ab_cross_ac(a,b,c),);
d2=dblcmp(s2=ab_cross_ac(a,b,d),);
d3=dblcmp(s3=ab_cross_ac(c,d,a),);
d4=dblcmp(s4=ab_cross_ac(c,d,b),); //如果规范相交则求交点
if ((d1^d2)==- && (d3^d4)==-)
{
p.x=(c.x*s2-d.x*s1)/(s2-s1);
p.y=(c.y*s2-d.y*s1)/(s2-s1);
return ;
} //如果不规范相交
if (d1== && point_on_line(c,a,b)<=)
{
p=c;
return ;
}
if (d2== && point_on_line(d,a,b)<=)
{
p=d;
return ;
}
if (d3== && point_on_line(a,c,d)<=)
{
p=a;
return ;
}
if (d4== && point_on_line(b,c,d)<=)
{
p=b;
return ;
}
//如果不相交
return -;
}
int main()
{
int t;
while(scanf("%d", &t)&&t)
{
int cnt=;
for(int i=;i<=t;i++)
scanf("%lf%lf%lf%lf", &line[i].a, &line[i].a.y, &line[i].b.x, &line[i].b.y);
for(int i=;i<=t;i++)
{
for(int j=i+;j<=t;j++)
{
if(ab_cross_cd(line[i].a, line[i].b, line[j].a, line[j].b)!=-)
cnt++;
}
}
printf("%d\n",cnt);
}
return ;
}
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