bzoj1857 [Scoi2010]传送带 [三分套三分]

Description

在一个2维平面上有两条传送带，每一条传送带可以看成是一条线段。两条传送带分别为线段AB和线段CD。lxhgww在AB上的移动速度为P，在CD上的移动速度为Q，在平面上的移动速度R。现在lxhgww想从A点走到D点，他想知道最少需要走多长时间

Input

输入数据第一行是4个整数，表示A和B的坐标，分别为Ax，Ay，Bx，By 第二行是4个整数，表示C和D的坐标，分别为Cx，Cy，Dx，Dy 第三行是3个整数，分别是P，Q，R

Output

输出数据为一行，表示lxhgww从A点走到D点的最短时间，保留到小数点后2位

Sample Input

0 0 0 100

100 0 100 100

2 2 1

Sample Output

136.60

HINT

对于100%的数据，1<= Ax，Ay，Bx，By，Cx，Cy，Dx，Dy<=1000

1<=P，Q，R<=10

Solution

三分在第一条传送带走多远，再三分第二条。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>

using namespace std;

const double pi = acos(-1);
const double eps = 1e-8;

double ax, ay, bx, by, cx, cy, dx, dy;
double p, q, r;
double dist(double x1, double y1, double x2, double y2) {
return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}

double cal(double x1, double y1, double x2, double y2) {
double d1 = dist(ax, ay, x1, y1);
double d2 = dist(x1, y1, x2, y2);
double d3 = dist(x2, y2, dx, dy);
return d1 / p + d2 / r + d3 / q;
}

double solve(double x, double y) {
double lx = cx, rx = dx;
double ly = cy, ry = dy;
while (fabs(rx - lx) > eps || fabs(ry - ly) > eps) {
double x1 = lx + (rx - lx) / 3.0;
double x2 = rx - (rx - lx) / 3.0;
double y1 = ly + (ry - ly) / 3.0;
double y2 = ry - (ry - ly) / 3.0;
double v1 = cal(x, y, x1, y1);
double v2 = cal(x, y, x2, y2);
if (v1 > v2) lx = x1, ly = y1;
else rx = x2, ry = y2;
}
return cal(x, y, lx, ly);
}

int main() {
// freopen("1857.in", "r", stdin);
scanf("%lf %lf %lf %lf", &ax, &ay, &bx, &by);
scanf("%lf %lf %lf %lf", &cx, &cy, &dx, &dy);
scanf("%lf %lf %lf", &p, &q, &r);
double lx = ax, rx = bx;
double ly = ay, ry = by;
while (fabs(rx - lx) > eps || fabs(ry - ly) > eps) {
double x1 = lx + (rx - lx) / 3.0;
double x2 = rx - (rx - lx) / 3.0;
double y1 = ly + (ry - ly) / 3.0;
double y2 = ry - (ry - ly) / 3.0;
double v1 = solve(x1, y1);
double v2 = solve(x2, y2);
if (v1 > v2) lx = x1, ly = y1;
else rx = x2, ry = y2;
}
printf("%.2lf\n", solve(lx, ly));
return 0;
}