題意：有一光源，被若干個圓摭住了，求地面上陰影
題解：利用向量旋轉求出切線與圓的兩個交點，根據兩點成線得出aX+bY+c=0的直線，令Y=0時，求出地面的坐標，再合並有連接的陰影。
//代碼如下：
[cpp]
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<math.h>
#define eps 1e-8
//using namespace std;
const int maxn=1000;
struct point{double x,y;};
struct circle {point cen;double r;}cir[maxn];
struct Node{double s,e;}node[maxn];
double f[maxn][2];
point st;
double distance(point p1,point p2)
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
void getline(point p1,point p2,double &a,double &b,double &c)
{
a=p2.y-p1.y;
b=p1.x-p2.x;
c=p2.x*p1.y-p1.x*p2.y;
}
void cirinser(circle c,point &inser1,point &inser2)//圓與切線交點(保證st在圓外)
{
point dir;
double dis=distance(c.cen,st);
double temp=sqrt(dis*dis-c.r*c.r);
double sina=temp/dis;
double cosa=c.r/dis;
dir.x=(st.x-c.cen.x)/dis*c.r;
dir.y=(st.y-c.cen.y)/dis*c.r;
inser1.x=c.cen.x+(dir.x*cosa-dir.y*sina);
inser1.y=c.cen.y+(dir.x*sina+dir.y*cosa);
inser2.x=c.cen.x+(dir.x*cosa+dir.y*sina);
inser2.y=c.cen.y+(-dir.x*sina+dir.y*cosa);
}
bool zero(double x){return x>0? x<eps:x>-eps;}
bool cmp( Node a, Node b)
{
if(zero(a.s-b.s))return a.e-b.e<0;
else return a.s-b.s<0;
}
int main()
{
int n;
while(scanf("%d",&n),n)
{
scanf("%lf%lf",&st.x,&st.y);
int i;
point inser1,inser2;
double a,b,c;
for(i=0;i<n;i++)
{
scanf("%lf%lf%lf",&cir[i].cen.x,&cir[i].cen.y,&cir[i].r);
cirinser(cir[i],inser1,inser2);
getline(st,inser1,a,b,c);
node[i].s=-c/a;
getline(st,inser2,a,b,c);
node[i].e=-c/a;
if(node[i].s-node[i].e>0)std::swap(node[i].s,node[i].e);
}
std::sort(node,node+n,cmp);
f[0][0]=node[0].s;
f[0][1]=node[0].e;
int cnt=0; www.2cto.com
for(i=1;i<n;i++)
{
if(node[i].s-f[cnt][1]<eps &&node[i].e-f[cnt][1]>0)f[cnt][1]=node[i].e;
else if(node[i].s-f[cnt][1]>0)
{
f[++cnt][0]=node[i].s;
f[cnt][1]=node[i].e;
}
}
for(i=0;i<=cnt;i++)
printf("%.2lf %.2lf\n",f[i][0],f[i][1]);
printf("\n");
}
return 0;
}
作者：ssslpk
