列出不等式后,把同时除Z把它去掉。
注意了,这里应该 是把直线变两点表示的向量更为简单,我开始就直接用直线写,后来,唉,写不下去了。。
#include <iostream> #include <algorithm> #include <cstdio> #include <cmath> using namespace std; const int MAX=110; const double eps=1e-10; const double inf=1e10; struct point{ double x,y; }; point operator -(point x,point y){ point t; t.x=x.x-y.x; t.y=x.y-y.y; return t; } double operator *(point x,point y){ return x.x*y.y-x.y*y.x; } struct line{ point a,b; double angle; }; int n,lnum; double A[MAX],B[MAX],C[MAX]; line le[MAX],st[MAX]; int DB(double d){ if(fabs(d)<eps) return 0; if(d>0) return 1; return -1; } void addLine(double x1,double y1,double x2,double y2){ point a; a.x=x1; a.y=y1; point b; b.x=x2; b.y=y2; le[lnum].a=a; le[lnum].b=b; le[lnum].angle=atan2(y2-y1,x2-x1); } bool cmp( line x, line y){ if(fabs(x.angle-y.angle)<eps){ if((y.b-x.a)*(x.b-y.a)>eps) return true; return false; } return x.angle<y.angle; } void getIntersect(line l1, line l2, point& p) { double dot1,dot2; dot1 = (l1.b-l2.a)*(l1.a-l2.a); dot2 = (l2.b-l1.b)*( l1.a-l1.b); p.x = (l2.a.x * dot2 + l2.b.x * dot1) / (dot2 + dot1); p.y = (l2.a.y * dot2 + l2.b.y * dot1) / (dot2 + dot1); } bool judge(line l0, line l1, line l2) { point p; getIntersect(l1, l2, p); return DB((l0.a-p)*(l0.b-p)) <=0; } bool pare(line x,line y){ return fabs((x.b-x.a)*(y.b-y.a))<eps; } bool half(){ int top=1,bot=0; sort(le,le+lnum,cmp); int tmp=1; for(int i=1;i<lnum;i++){ if(fabs(le[i].angle-le[tmp-1].angle)>eps) le[tmp++]=le[i]; } lnum=tmp; st[0]=le[0]; st[1]=le[1]; for(int i=2;i<lnum;i++){ while(bot<top&&judge(le[i], st[top-1], st[top])) top--; while(bot<top&&judge(le[i],st[bot+1],st[bot])) bot++; st[++top]=le[i]; } while(bot<top&&judge(st[bot],st[top-1],st[top])) top--; while(bot<top&&judge(st[top],st[bot+1],st[bot])) bot++; if(top<=bot+1) return false; return true; } bool slove(int s){ int i, j, k; double x1, y1, x2, y2, a, b, c; lnum=0; addLine(0, 0, inf, 0 ); lnum++; addLine(inf, 0, inf, inf ); lnum++; addLine(inf, inf, 0, inf ); lnum++; addLine(0, inf, 0, 0 ); lnum++; for (j = 0; j < n; j++) if (s != j) { a = 1.0 / A[j] - 1.0 / A[s]; b = 1.0 / B[j] - 1.0 / B[s]; c = 1.0 / C[j] - 1.0 / C[s]; int d1 = DB(a); int d2 = DB(b); int d3 = DB(c); if (!d1) { if (!d2) { if (d3 <= 0) { return false; } continue; } x1 = 0; x2 = d2; y1 = y2 = - c / b; } else { if (!d2) { x1 = x2 = - c / a; y1 = 0; y2 = -d1; } else { x1 = 0; y1 = - c / b; x2 = d2; y2 = -(c + a * x2) / b; } } addLine(x1, y1, x2, y2 ); lnum++; } if(!half()) return false; return true; } int main(){ while(scanf("%d",&n)!=EOF){ for(int i=0;i<n;i++) scanf("%lf%lf%lf",&A[i],&B[i],&C[i]); for(int i=0;i<n;i++){ if(slove(i)) printf("Yes\n"); else printf("No\n"); } } return 0; }
POJ 1755
时间: 2024-11-04 11:58:48