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hdu 1086线段相交模板

2013-12-12 00:00 267 查看
考虑情况要全面
http://acm.hdu.edu.cn/showproblem.php?pid=1086
code:

#include <iostream>
#include "cstdio"
#include "cmath"
#include "algorithm"
using namespace std;
struct point
{
double x,y;
};
point a[105][2];
double fan(double x,double y)
{
return x>y?x:y;
}
double fin(double c,double d)
{
return c<d?c:d;
}
double cnt(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int is(point a,point b,point c,point d)	//线段相交模板
{
if(a.x==b.x&&c.x==d.x)	//两条线重合了
{
return 0;
}
if(a.x==b.x&&c.x!=d.x)	//第一条线平行于y轴
{
double m1=a.x;
double m2=(a.x-c.x)*(d.y-c.y)/(d.x-c.x)+c.y;
if(m1<=fan(a.x,b.x)&&m1>=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&(m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y)))
{
return 1;
}
}
if(c.x==d.x&&a.x!=b.x)	//第二条线平行y轴
{
double m1=c.x;
double m2=a.y+(b.y-a.y)*(c.x-a.x)/(b.x-a.x);
if(m1<=fan(a.x,b.x)&&m1>=fin(a.x,b.x)&&m2>=fin(a.y,b.y)&&m2<=fan(a.y,b.y)&&(m1<=fan(c.x,d.x)&&m1>=fin(c.x,d.x)&&m2>=fin(c.y,d.y)&&m2<=fan(c.y,d.y)))
{
return 1;
}
}
double k1=(b.y-a.y)/(b.x-a.x);
double k2=(d.y-c.y)/(d.x-c.x);
double m1,m2,x,y;
if(k1==k2)	//斜率相同即平行
{
return 0;
}else
{
m1=a.y-k1*a.x;
m2=c.y-k2*c.x;
x=(m1-m2)/(k2-k1);
y=k1*x+m1;
if(x<=fan(a.x,b.x)&&x>=fin(a.x,b.x)&&y>=fin(a.y,b.y)&&y<=fan(a.y,b.y)&&(x<=fan(c.x,d.x)&&x>=fin(c.x,d.x)&&y>=fin(c.y,d.y)&&y<=fan(c.y,d.y)))
{
return 1;
}
}
return 0;

}
int main(int argc, char *argv[])
{
int cas=1;
int n,i,j;
while(scanf("%d",&n)&&n)
{
int cnt=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf%lf%lf",&a[i][0].x,&a[i][0].y,&a[i][1].x,&a[i][1].y);	//先输x,y起点再输x,y终点
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(is(a[i][0],a[i][1],a[j][0],a[j][1]))
{
cnt+=1;
}
}
}
printf("%d\n",cnt);
}

return 0;
}
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