CSU 1412 Line and Circles
时间:2014-04-27 21:00:32
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原题链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1412
题目要求判断是否有一条直线可以穿过所有的圆。
做法:把所有圆心做一次凸包,然后判断这个凸包是否能通过一个宽度为2*R的通道。
做法和求凸包直径差不多,只是判断的时候把点到两个端点的距离换成点到直线的距离。
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <algorithm> using namespace std; #define inf 1e20 #define eps 1e-8 const int N = 100005 ; const double PI = 2.0*asin(1.0); //高精度求PI struct Lpoint { double x,y; }a[N], b[N]; //点 struct Llineseg { Lpoint a,b; }; //线段 struct Ldir { double dx,dy; }; //方向向量 struct Lline { Lpoint p; Ldir dir; }; //直线 bool mult(Lpoint sp, Lpoint ep, Lpoint op) { return (sp.x - op.x) * (ep.y - op.y) >= (ep.x - op.x) * (sp.y - op.y); } bool operator < (const Lpoint &l, const Lpoint &r) { return l.y < r.y || (l.y == r.y && l.x < r.x); } int graham(Lpoint pnt[], int n, Lpoint res[]) { int i, len, top = 1; sort(pnt, pnt + n); if (n == 0) return 0; res[0] = pnt[0]; if (n == 1) return 1; res[1] = pnt[1]; if (n == 2) return 2; res[2] = pnt[2]; for (i = 2; i < n; i++) { while (top && mult(pnt[i], res[top], res[top-1])) top--; res[++top] = pnt[i]; } len = top; res[++top] = pnt[n - 2]; for (i = n - 3; i >= 0; i--) { while (top!=len && mult(pnt[i], res[top], res[top-1])) top--; res[++top] = pnt[i]; } return top; // 返回凸包中点的个数 } void format(Lline ln, double& A, double& B, double& C) { A=ln.dir.dy; B=-ln.dir.dx; C=ln.p.y*ln.dir.dx-ln.p.x*ln.dir.dy; } double p2ldis(Lpoint a, Lline ln) { double A,B,C; format(ln,A,B,C); return(fabs(A*a.x+B*a.y+C)/sqrt(A*A+B*B)); } double CPMD(Lpoint p[], int n)//ConvexPolygonMinimumDiameter { int i, j; double ans = inf, tmp; p[n] = p[0]; Lline ln; Ldir dir; for(i = 0, j = 1; i < n; ++i) { if((i+1)%n == j) j = (j + 1) % n; dir.dx = p[i].x - p[i+1].x; dir.dy = p[i].y - p[i+1].y; ln.dir = dir; ln.p = p[i]; while((tmp = p2ldis(p[j], ln)) < (p2ldis(p[(j+1)%n], ln))) j = (j + 1) % n; ans = min(ans, tmp); } return ans; } double dis(Lpoint u, Lpoint v) { return sqrt((u.x-v.x) * (u.x-v.x) + (u.y - v.y)*(u.y - v.y)); } int main() { int n, t; double r; scanf("%d", &t); while(t--) { scanf("%d%lf", &n, &r); for(int i = 0; i < n; i++) scanf("%lf%lf", &a[i].x, &a[i].y); int m = graham(a, n, b); if(m <= 2) { printf("Yes\n"); continue; } double k = CPMD(b, m); if(k - 2*r < eps) printf("Yes\n"); else printf("No\n"); } return 0; }
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