UVA 10566 - Crossed Ladders(二分+计算几何)

这个很显然，交点高度和底边长度成反比例函数，可以用二分求解

二分底边，在利用交点求出高度，判断即可

代码：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;struct Point {    double x, y;    Point() {}    Point(double x, double y) {        this->x = x;        this->y = y;    }    void read() {        scanf("%lf%lf", &x, &y);    }};typedef Point Vector;Vector operator + (Vector A, Vector B) {    return Vector(A.x + B.x, A.y + B.y);}Vector operator - (Vector A, Vector B) {    return Vector(A.x - B.x, A.y - B.y);}Vector operator * (Vector A, double p) {    return Vector(A.x * p, A.y * p);}Vector operator / (Vector A, double p) {    return Vector(A.x / p, A.y / p);}double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {    Vector u = P - Q;    double t = Cross(w, u) / Cross(v, w);    return P + v * t;}double x, y, C;int main() {    while (~scanf("%lf%lf%lf", &x, &y, &C)) {        double l = 0, r = min(x, y);        for (int i = 0; i < 100; i++) {            double mid = (l + r) / 2;            Point a = Point(0, 0);            Point b = Point(mid, sqrt(y * y - mid * mid));            Point c = Point(0, sqrt(x * x - mid * mid));            Point d = Point(mid, 0);            double h = GetLineIntersection(a, b - a, c, d - c).y;            if (h > C) l = mid;            else r = mid;        }        printf("%.3f\n", l);    }    return 0;}