发现当把 \(\rm scale×cos\theta,scale×sin\theta,dx,dy\) 当作变量时只有四个,两个方程就行。
当 \(\rm n\le 500\) 时,可以选取两组进行高斯消元,解出答案后回带。
但当 \(n\) 极大时,采用随机化的做法,每次随机选取两个,这样每次选取不正确的概率为 \(\frac{3}{4}\),选取 50 次后基本就会出答案了。
记得判断 \(\rm sin\) 的正负
Code
#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf;
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
struct nanfeng_stream{
template<typename T>inline nanfeng_stream &operator>>(T &x) {
ri f=0;x=0;register char ch=gc();
while(!isdigit(ch)) {f|=ch=='-';ch=gc();}
while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
return x=f?-x:x,*this;
}
}cin;
}
using IO::cin;
namespace nanfeng{
#define FI FILE *IN
#define FO FILE *OUT
template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
typedef double db;
static const int N=1e5+7;
static const db eps=1e-6;
struct node{db x1,y1,x2,y2;}pnt[N];
int p[N],n,hl;
db mt[51][51],x1,x2,x3,x4;
inline void Guass() {
for (ri i(1);i<=4;p(i)) {
ri k=i;
for (ri j(i+1);j<=4;p(j)) if (fabs(mt[j][i])>fabs(mt[k][i])) k=j;
if (k!=i) std::swap(mt[k],mt[i]);
for (ri j(1);j<=4;p(j)) {
if (i==j) continue;
db tmp=mt[j][i]/mt[i][i];
for (ri l(1);l<=5;p(l)) mt[j][l]-=tmp*mt[i][l];
}
}
}
inline bool judge() {
ri cnt(0);
x1=mt[1][5]/mt[1][1];
x2=mt[2][5]/mt[2][2];
x3=mt[3][5]/mt[3][3];
x4=mt[4][5]/mt[4][4];
for (ri i(1);i<=n;p(i))
if (fabs(pnt[i].x1*x1-pnt[i].y1*x2+x3-pnt[i].x2)<=eps&&fabs(pnt[i].x1*x2+pnt[i].y1*x1+x4-pnt[i].y2)<=eps) p(cnt);
return cnt>=hl;
}
inline int solve() {
register db sc=sqrt(x1*x1+x2*x2);
if (x2/sc<=eps) printf("%.10lf\n",acos(-1)*2.0-acos(x1/sc));
else printf("%.10lf\n",acos(x1/sc));
printf("%.10lf\n%.10lf %.10lf\n",sc,x3,x4);
return 0;
}
inline int main() {
//FI=freopen("nanfeng.in","r",stdin);
//FO=freopen("nanfeng.out","w",stdout);
srand(unsigned(time(0)));
std::cin >> n;
hl=(n>>1)+(n&1);
for (ri i(1);i<=n;p(i)) {
scanf("%lf%lf%lf%lf",&pnt[i].x1,&pnt[i].y1,&pnt[i].x2,&pnt[i].y2);
p[i]=i;
}
if (n>500) {
std::random_shuffle(p+1,p+n+1);
for (ri i(1);i<=n;p(i)) {
ri cur1(p[i]),cur2(p[i+1]);
mt[1][1]=pnt[cur1].x1,mt[1][2]=-pnt[cur1].y1,mt[1][3]=1.0,mt[1][4]=0.0,mt[1][5]=pnt[cur1].x2;
mt[2][1]=pnt[cur1].y1,mt[2][2]=pnt[cur1].x1,mt[2][3]=0.0,mt[2][4]=1.0,mt[2][5]=pnt[cur1].y2;
mt[3][1]=pnt[cur2].x1,mt[3][2]=-pnt[cur2].y1,mt[3][3]=1.0,mt[3][4]=0.0,mt[3][5]=pnt[cur2].x2;
mt[4][1]=pnt[cur2].y1,mt[4][2]=pnt[cur2].x1,mt[4][3]=0.0,mt[4][4]=1.0,mt[4][5]=pnt[cur2].y2;
Guass();
if (judge()) return solve();
}
} else {
for (ri i(1);i<=n;p(i))
for (ri j(i+1);j<=n;p(j)) {
ri cur1(i),cur2(j);
mt[1][1]=pnt[cur1].x1,mt[1][2]=-pnt[cur1].y1,mt[1][3]=1.0,mt[1][4]=0.0,mt[1][5]=pnt[cur1].x2;
mt[2][1]=pnt[cur1].y1,mt[2][2]=pnt[cur1].x1,mt[2][3]=0.0,mt[2][4]=1.0,mt[2][5]=pnt[cur1].y2;
mt[3][1]=pnt[cur2].x1,mt[3][2]=-pnt[cur2].y1,mt[3][3]=1.0,mt[3][4]=0.0,mt[3][5]=pnt[cur2].x2;
mt[4][1]=pnt[cur2].y1,mt[4][2]=pnt[cur2].x1,mt[4][3]=0.0,mt[4][4]=1.0,mt[4][5]=pnt[cur2].y2;
Guass();
if (judge()) return solve();
}
}
return 0;
}
}
int main() {return nanfeng::main();}
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