标签：bool i++ log 最大的 end can clu hoj 计算
现在给出一个椭球面: $ax ^ 2 + by ^ 2 + cz ^ 2 + dyz + exz + fxy = 1$ .
求椭球面到 $(0, 0, 0)$ 的距离.
$T \le 200, 0 < a, b, c < 1, 0 \le d, e, f < 1$ .
假装 $(0, 0, 0)$ 在椭球内部.
二次的式子通常都是单峰的.
猜测椭球到 $(0, 0, 0)$ 的距离也是单峰的.
三分 x , 三分 y , 利用一元二次方程解出 z 并计算距离.
实现小结:
1. 对于三分, 模拟退火之类的, 最好记录访问到的最大的答案, 而不是最后再求一次.
2. 浮点数上, 如果运算出错, 会返回 nan , 判断一个数是不是 nan , 就判断 !(x < 0) && !(x >= 0) .
最终实现:
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cctype> 5 #include <cmath> 6 #include <algorithm> 7 using namespace std; 8 #define F(i, a, b) for (register int i = (a); i <= (b); i++) 9 #define db double 10 11 const db INF = 1e30; 12 13 db a, b, c, d, e, f, ans; 14 15 inline bool nan(db x) { return !(x < 0) && !(x >= 0); } 16 inline db Dist(db x, db y, db z) { return sqrt(x * x + y * y + z * z); } 17 inline db Calc(db x, db y) { 18 db A = c; 19 db B = d * y + e * x; 20 db C = a * x * x + b * y * y + f * x * y - 1; 21 db D = B * B - 4 * A * C; 22 db z = (-B - sqrt(D)) / (2 * A); 23 db tmp = Dist(x, y, z); 24 return ans = min(ans, tmp), tmp; 25 } 26 inline db Y(db x) { 27 db l = 0, r = 1000 / sqrt(b); 28 while (r-l > 1e-10) { 29 db _l = (l+l+r)/3, ansL = Calc(x, _l); 30 db _r = (l+r+r)/3, ansR = Calc(x, _r); 31 nan(ansL) || nan(ansR) || ansL < ansR ? r = _r : l = _l; 32 } 33 db ans = Calc(x, l); 34 35 l = -1000 / sqrt(b), r = 0; 36 while (r-l > 1e-10) { 37 db _l = (l+l+r)/3, ansL = Calc(x, _l); 38 db _r = (l+r+r)/3, ansR = Calc(x, _r); 39 nan(ansL) || nan(ansR) || ansR < ansL ? l = _l : r = _r; 40 } 41 db ans2 = Calc(x, l); 42 43 return nan(ans) ? ans2 : nan(ans2) ? ans : min(ans, ans2); 44 } 45 inline db X(void) { 46 db l = 0, r = 1000 / sqrt(a); 47 while (r-l > 1e-10) { 48 db _l = (l+l+r)/3, ansL = Y(_l); 49 db _r = (l+r+r)/3, ansR = Y(_r); 50 nan(ansL) || nan(ansR) || ansL < ansR ? r = _r : l = _l; 51 } 52 db ans = Y(l); 53 54 l = -1000 / sqrt(a), r = 0; 55 while (r-l > 1e-10) { 56 db _l = (l+l+r)/3, ansL = Y(_l); 57 db _r = (l+r+r)/3, ansR = Y(_r); 58 nan(ansL) || nan(ansR) || ansR < ansL ? l = _l : r = _r; 59 } 60 db ans2 = Y(l); 61 62 return nan(ans) ? ans2 : nan(ans2) ? ans : min(ans, ans2); 63 } 64 65 int main(void) { 66 #ifndef ONLINE_JUDGE 67 freopen("e.in", "r", stdin); 68 #endif 69 70 while (~scanf("%lf %lf %lf %lf %lf %lf", &a, &b, &c, &d, &e, &f)) 71 ans = INF, X(), printf("%0.8lf\n", ans); 72 73 return 0; 74 }
标签：bool i++ log 最大的 end can clu hoj 计算
原文地址：http://www.cnblogs.com/Sdchr/p/7505582.html
