pid=4454" target="_blank" style="">题目链接:hdu 4454 Stealing a Cake
题目大意:给定一个起始点s,一个圆形。一个矩形。如今从起点開始,移动到圆形再移动到矩形。求最短距离。
解题思路:在圆周上三分就可以。即对角度[0,2*pi]三分。计算点和矩形的距离能够选点和矩形四条边的距离最短值。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-9;
const double pi = 4 * atan(1.0);
struct point {
double x, y;
point(double x = 0, double y = 0) {
this->x = x;
this->y = y;
}
}s, o, p[4];
double R;
inline double distant(point u, point v) {
double x = u.x - v.x;
double y = u.y - v.y;
return sqrt(x*x + y*y);
}
inline double handle(point u, point l, point r) {
if (fabs(l.x - r.x) < eps) {
double a = min(l.y, r.y), b = max(l.y, r.y);
if (a <= u.y && u.y <= b)
return fabs(u.x - l.x);
else
return min(distant(u, l), distant(u, r));
} else {
double a = min(l.x, r.x), b = max(l.x, r.x);
if (a <= u.x && u.x <= b)
return fabs(u.y - l.y);
else
return min(distant(u, l), distant(u, r));
}
}
inline double f(double k) {
point u(o.x + R * cos(k), o.y + R * sin(k));
double ret = handle(u, p[0], p[3]);
for (int i = 0; i < 3; i++)
ret = min(ret, handle(u, p[i], p[i+1]));
return distant(s, u) + ret;
}
double solve (double l, double r) {
for (int i = 0; i < 100; i++) {
double x1 = l + (r-l) / 3;
double x2 = r - (r-l) / 3;
if (f(x1) < f(x2))
r = x2;
else
l = x1;
}
return f(l);
}
void init () {
double a, b, c, d;
scanf("%lf%lf%lf", &o.x, &o.y, &R);
scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
p[0].x = min(a, c); p[0].y = min(b, d);
p[1].x = min(a, c); p[1].y = max(b, d);
p[2].x = max(a, c); p[2].y = max(b, d);
p[3].x = max(a, c); p[3].y = min(b, d);
/*
for (int i = 0; i < 4; i++)
printf("%lf %lf\n", p[i].x, p[i].y);
*/
}
int main () {
while (scanf("%lf%lf", &s.x, &s.y) == 2 && (fabs(s.x) > eps || fabs(s.y) > eps)) {
init();
printf("%.2lf\n", solve(0, 2 * pi));
}
return 0;
}
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