NOIP 模擬 $32\; \rm Walker$
阿新 • • 發佈:2021-08-07
題解
題解 \(by\;zj\varphi\)
發現當把 \(\rm scale×cos\theta,scale×sin\theta,dx,dy\) 當作變數時只有四個,兩個方程就行。
當 \(\rm n\le 500\) 時,可以選取兩組進行高斯消元,解出答案後迴帶。
但當 \(n\) 極大時,採用隨機化的做法,每次隨機選取兩個,這樣每次選取不正確的概率為 \(\frac{3}{4}\),選取 50 次後基本就會出答案了。
記得判斷 \(\rm sin\) 的正負
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();}