Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 2005 Solved: 1091
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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
Source
Day2
很显然最优的路线一定是从A走到AB上的一点走到CD上的一点再走到D
然后三分套三分就可以了
代码语言：javascript复制#include<cstdio>
#include<cmath>
#include<algorithm>
#define eps 1e-3
using namespace std;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 c - '0',c = getchar();
return x * f;
}
int Ax, Ay, Bx, By, Cx, Cy, Dx, Dy, P, Q, R;
double dis(double x, double y, double x2, double y2) {
return sqrt((x2 - x) * (x2 -x) (y2 - y) * (y2 - y));
}
double check2(double x,double y,double x2,double y2) {
return dis(Ax, Ay, x, y) / P dis(x, y, x2, y2) / R dis(x2, y2, Dx, Dy) / Q;
}
double check(double x, double y) {
double lx = Cx, ly = Cy, rx = Dx, ry = Dy, a, b;
while(abs(rx - lx) > eps || abs(ry - ly) > eps) {
double wx1 = (lx * 2 rx) / 3, wy1 = (ly * 2 ry) / 3,
wx2 = (lx rx * 2) / 3, wy2 = (ly ry * 2) / 3;
a = check2(x, y, wx1, wy1);
b = check2(x, y, wx2, wy2);
if(a > b) lx = wx1, ly = wy1;
else rx = wx2, ry = wy2;
}
return check2(x, y, lx, ly);
}
int main() {
#ifdef WIN32
freopen("a.in","r",stdin);
#endif
Ax = read(), Ay = read(), Bx = read(), By = read(), Cx = read(), Cy = read(), Dx = read(), Dy = read(), P = read(), Q = read(), R = read();
double lx = Ax, ly = Ay, rx = Bx, ry = By;
while(abs(rx - lx) > eps || abs(ry - ly) > eps) {
double wx1 = (lx * 2 rx) / 3, wy1 = (ly * 2 ry) / 3,
wx2 = (lx rx * 2) / 3, wy2 = (ly ry * 2) / 3;
if(check(wx1, wy1) > check(wx2, wy2)) lx = wx1, ly = wy1;
else rx = wx2, ry = wy2;
}
printf("%.2lf", check(lx, ly));
return 0;
}
