标签:oid with wap return circle class cin lag --
http://exam.upc.edu.cn/problem.php?id=5252
斜截式表示的直线方程
求三角形的内切圆和外接圆
求直线与直线交点,直线与圆交点
1 #include <bits/stdc++.h> 2 using namespace std; 3 const long double EPS = 0.0000001; 4 5 // a == b return 0 6 // a > b return 1 7 // a < b return -1 8 inline int cmp(long double a, long double b) { 9 if(abs(a - b) < EPS) return 0; 10 return a > b? 1: -1; 11 } 12 struct point { 13 long double x, y; 14 void print() { 15 printf(" (%.4Lf, %.4Lf)\n", x, y); 16 } 17 }; 18 struct line { 19 long double k, b; 20 bool isk = true; 21 line() {} 22 line(long double x1, long double y1, long double x2, long double y2) { 23 if(!cmp(x1, x2)) { 24 isk = false; 25 b = x1; 26 } else { 27 isk = true; 28 k = (y1 - y2) / (x1 - x2); 29 b = y1 - k * x1; 30 } 31 } 32 line(point a, point b) { 33 *this = line(a.x, a.y, b.x, b.y); 34 } 35 void print() { 36 printf(" k = %.4Lf, b = %.4Lf\n", k, b); 37 } 38 }; 39 struct circle { 40 point o; 41 long double r; 42 void print() { 43 printf(" center:(%.4Lf, %.4Lf) rad: %.4Lf\n", o.x, o.y, r); 44 } 45 }; 46 inline long double dist (point a, point b) { 47 return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); 48 } 49 circle getWjc(point a, point b, point c) { 50 long double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/2; 51 long double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/2; 52 long double d = a1 * b2 - a2 * b1; 53 circle cir; 54 cir.o.x = a.x + (c1*b2 - c2*b1)/d; 55 cir.o.y = a.y + (a1*c2 - a2*c1)/d; 56 cir.r = dist(cir.o, a); 57 return cir; 58 } 59 circle getNqc(point a, point b, point c) { 60 long double C = dist(a, b); 61 long double B = dist(a, c); 62 long double A = dist(b, c); 63 circle cir; 64 cir.o.x = (A*a.x + B*b.x + C*c.x) / (A + B + C); 65 cir.o.y = (A*a.y + B*b.y + C*c.y) / (A + B + C); 66 cir.r = sqrt((A + B - C)*(A - B + C)*(-A + B + C) / (A + B + C)) / 2; 67 return cir; 68 } 69 pair<point, point> getPoint(circle cir, line l) { 70 pair<point, point> ans; 71 if(l.isk) { 72 long double x0 = cir.o.x; 73 long double y0 = cir.o.y; 74 long double a = (1 + l.k * l.k); 75 long double b = (2*l.k*(l.b-y0)-2*x0); 76 long double c = (x0*x0+(l.b-y0)*(l.b-y0)-cir.r*cir.r); 77 long double deta = b * b - 4 * a * c; 78 long double x1 = ( - b - sqrt(deta)) / (2 * a); 79 long double x2 = ( - b + sqrt(deta)) / (2 * a); 80 long double y1 = l.k*x1+l.b; 81 long double y2 = l.k*x2+l.b; 82 ans.first.x = x1; 83 ans.first.y = y1; 84 ans.second.x = x2; 85 ans.second.y = y2; 86 } else { 87 long double x0 = cir.o.x; 88 long double y0 = cir.o.y; 89 long double y1 = y0 - sqrt(cir.r*cir.r - (l.b - x0) * (l.b - x0)); 90 long double y2 = y0 + sqrt(cir.r*cir.r - (l.b - x0) * (l.b - x0)); 91 ans.first.x = l.b; 92 ans.first.y = y1; 93 ans.second.x = l.b; 94 ans.second.y = y2; 95 } 96 return ans; 97 } 98 point getPoint(line l1, line l2) { 99 point ans; 100 if( ((!l1.isk) && (!l2.isk)) || (l1.isk && l2.isk && !cmp(l1.k, l2.k)) ) { 101 cerr << "RONG!!!!" << endl; exit(-1); 102 } else { 103 if(!l2.isk) swap(l1, l2); 104 if(l1.isk) { 105 long double b1 = l1.b; 106 long double b2 = l2.b; 107 long double k1 = l1.k; 108 long double k2 = l2.k; 109 long double x = (b2 - b1) / (k1 - k2); 110 long double y = k1*x+b1; 111 ans.x = x; 112 ans.y = y; 113 } else { 114 long double x = l1.b; 115 long double y = l2.k * x + l2.b; 116 ans.x = x; 117 ans.y = y; 118 } 119 } 120 return ans; 121 } 122 point myGetPoint(circle cir, line l, point p) { 123 pair<point, point> points = getPoint(cir, l); 124 long double d1 = dist(points.first, p); 125 long double d2 = dist(points.second, p); 126 if(d1 > d2) return points.first; 127 else return points.second; 128 } 129 int main() 130 { 131 //freopen("in.txt", "r", stdin); 132 ios::sync_with_stdio(false); 133 int p; cin >> p; 134 while(p--) { 135 int k; cin >> k; cout << k << ‘ ‘; 136 long double bx, cx, cy; cin >> bx >> cx >> cy; 137 point A, B, C; 138 A.x = 0; A.y = 0; 139 B.x = bx; B.y = 0; 140 C.x = cx; C.y = cy; 141 circle nqy = getNqc(A, B, C); //nqy.print(); 142 circle wjy = getWjc(A, B, C); //wjy.print(); 143 point I = nqy.o; 144 line CP(C, I); 145 line AM(A, I); 146 line BN(B, I); 147 point P = myGetPoint(wjy, CP, C); 148 point M = myGetPoint(wjy, AM, A); 149 point N = myGetPoint(wjy, BN, B); 150 line PN(P, N); 151 line NM(N, M); 152 line MP(M, P); 153 line AB(A, B); 154 line BC(B, C); 155 line CA(C, A); 156 point E = getPoint(PN, AB); 157 point F = getPoint(CA, PN); 158 point G = getPoint(NM, CA); 159 point H = getPoint(NM, BC); 160 point J = getPoint(BC, MP); 161 point K = getPoint(AB, MP); 162 cout.flags(ios::fixed); 163 cout.precision(4); 164 cout << dist(E, F) << ‘ ‘; 165 cout << dist(F, G) << ‘ ‘; 166 cout << dist(G, H) << ‘ ‘; 167 cout << dist(H, J) << ‘ ‘; 168 cout << dist(J, K) << ‘ ‘; 169 cout << dist(K, E) << endl; 170 } 171 return 0; 172 } 173
计算几何:直线与圆的交点 三角形的内切圆和外接圆(5252: Triangle to Hexagon)
标签:oid with wap return circle class cin lag --
原文地址:https://www.cnblogs.com/upc201713/p/8978833.html