hdu1154(点是否在多边形内部、两线段交点）
point intersection(line &u,line &v)//两线段之间的交点
{
point ret=u.a;
double t=((u.a.x-v.a.x)*(v.a.y-v.b.y) - (u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
ret.x+=(u.b.x-u.a.x)*t;
ret.y+=(u.b.y-u.a.y)*t;
return ret;
}
题意是一条直线与多边形相交，求相交的总长，最后没怎么搞懂
#include
#include
#include
#include
#include
using namespace std;
#define maxn 1200
#define eps 1e-8
struct point
{
double x;
double y;
}po[maxn],rec[maxn];
struct line
{
point a;
point b;
}temp,temp1;
double xmulit(point &a,point &b,point &c)//差乘
{
return (a.x-b.x)*(a.y-c.y)-(a.y-b.y)*(a.x-c.x);
}
bool across(point &a,point &b,point &c,point &d)//直线ab和线段cd是否相交
{
double p=xmulit(a,b,c),p1=xmulit(a,b,d);
if( fabs(p1) <= eps || fabs(p) <= eps )//c或者d在ab上
return true;
if( p*p1 < -eps )//c和d在ab两侧，不用判断a和b是不是在cd两侧，因为ab是直线line
return true;
return false;
}
bool one_line(point &a,point &b,point &c,point &d)//直线ab和线段cd是否重合，相当于p==0&&p1==0
{
double p=xmulit(a,b,c),p1=xmulit(a,b,d);
if( fabs(p1) < eps && fabs(p) < eps )
return true;
return false;
}
bool is_equal(point &a,point &b)//判断点a和点b是否相等
{
return (fabs(a.x-b.x) <= eps) && (fabs(a.y-b.y) <=eps);
}
point intersection(line &u,line &v)//两线段之间的交点
{
point ret=u.a;
double t=((u.a.x-v.a.x)*(v.a.y-v.b.y) - (u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
ret.x+=(u.b.x-u.a.x)*t;
ret.y+=(u.b.y-u.a.y)*t;
return ret;
}
int n,m;
double dis(point &a,point &b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int cmp(point a, point b)
{
if(fabs(a.x-b.x)<eps)
return a.y
return a.x
}
bool on_segment(point pi,point pj,point pk)//判断点pk时候在线段pi, pj上
{
if(xmulit(pi, pj, pk)==0)
{
if(pk.x>=min(pi.x,pj.x)&&pk.x<=max(pi.x,pj.x)&&pk.y>=min(pi.y,pj.y)&&pk.y<=max(pi.y,pj.y))
return true;
}
return false;
}
bool segments_intersect(point p1,point p2,point p3,point p4)//判断线段是否相交
{
double d1=xmulit(p3,p4,p1);
double d2=xmulit(p3,p4,p2);
double d3=xmulit(p1,p2,p3);
double d4=xmulit(p1,p2,p4);
if(d1*d2<0&&d3*d4<0)
return true;
else if(d1==0&&on_segment(p3,p4,p1))
return true;
else if(d2==0&&on_segment(p3,p4,p2))
return true;
else if(d3==0&&on_segment(p1,p2,p3))
return true;
else if(d4==0&&on_segment(p1,p2,p4))
return true;
return false;
}
int inpoto(point a)//判断点是否在多边形的内部
{
int i;
point b,c,d;
b.y=a.y;
b.x=1e15;//定义射线
int count=0;
for(i=0;i<n;i++)
{
c = po[i];
d = po[i + 1];
if(on_segment(c,d,a))//该点在多边形的一条边上
return 1;
if(abs(c.y-d.y)<eps)
continue;
if(on_segment(a,b,c))//和顶点相交的情况，如果y值较大则取
{
if(c.y>d.y)
count++;
}
else if(on_segment(a,b,d))//和顶点相交的情况，如果y值较大则取
{
if(d.y>c.y)
count++;
}
else if(segments_intersect(a,b,c,d))//和边相交
count++;
}
return count%2;//当L和多边形的交点数目C是奇数的时候，P在多边形内，是偶数的话P在多边形外。
}
point mid(point &a,point &b)
{
point c;
c.x=(a.x+b.x)/2;
c.y=(a.y+b.y)/2;
return c;
}
double find_ans()
{
int i,pos=0;
double ans=0;
po[n]=po[0];
for(i=0;i<n;i++)
{
if(one_line(temp.a,temp.b,po[i],po[i+1]))//线段在直线上
{
rec[pos++]=po[i];
rec[pos++]=po[i+1];
//printf("pos=%d\n",pos);
continue;
}
if(across(temp.a,temp.b,po[i],po[i+1]))//线段与直线相交
{
temp1.a=po[i],temp1.b=po[i+1];
rec[pos++]=intersection(temp,temp1);//两条线的交点
}
}
for(int i=0;i
printf("%lf %lf\n",rec[i].x,rec[i].y);
printf("*****\n");
sort(rec,rec+pos,cmp);
for(int i=0;i
printf("%lf %lf\n",rec[i].x,rec[i].y);
printf("~~~~~\n");
for(i=0;i
{
if(inpoto(mid(rec[i],rec[i+1])))
ans+=dis(rec[i],rec[i+1]);
}
return ans;
}
int main()
{
int i;
while(scanf("%d %d",&n,&m))
{
if(m==0 && m==0)
return 0;
for(i=0;i<n;i++)
scanf("%lf%lf",&po[i].x,&po[i].y);
for(i=0;i<m;i++)
{
scanf("%lf%lf%lf%lf",&temp.a.x,&temp.a.y,&temp.b.x,&temp.b.y);
printf("%.3lf\n",find_ans());
}
}
return 0;
}