hdu 5134 Highway

其实是一个圆和六边形的面积并。

可以知道的是，如图1，对于高速公路上的y点，一定是如图红线的方式走的最远（即使得当时的x值最大）。

于是可以得到最终的区域即为图2的红色区域。

对于P点，我们可以通过三分得到，从起点到P的时间t，于是问题就得到了解决。圆和多边形面积并下就好了~

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<string>#include<queue>#include<cmath>#include<stack>#include<set>#include<map>#define FIR first#define SEC second#define MP make_pair#define inf 0x3f3f3f3f#define LL long long#define CLR(a, b) memset(a, b, sizeof(a))using namespace std;struct Point{    double x, y;    Point(double x = 0, double y = 0)        :x(x), y(y) {}};double EP = 0;double x_mult(Point sp, Point ep, Point op){    return (sp.x-op.x)*(ep.y-op.y)-(sp.y-op.y)*(ep.x-op.x);}double cross(Point a,Point b,Point c){    return (a.x-c.x)*(b.x-c.x)+(a.y-c.y)*(b.y-c.y);}double dist(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double cal_area(Point a,Point b,Point c,double r){    double A,B,C,x,y,tS;    A=dist(b,c);    B=dist(a,c);    C=dist(b,a);    if(A<r&&B<r)    return x_mult(a,b,c)/2;    else if(A<r&&B>=r){        x=(cross(a,c,b)+sqrt(r*r*C*C-x_mult(a,c,b)*x_mult(a,c,b)))/C;        tS=x_mult(a,b,c)/2;        return asin(tS*(1-x/C)*2/r/B*(1-EP))*r*r/2+tS*x/C;    }    else if(A>=r&&B<r){        y=(cross(b,c,a)+sqrt(r*r*C*C-x_mult(b,c,a)*x_mult(b,c,a)))/C;        tS=x_mult(a,b,c)/2;        return asin(tS*(1-y/C)*2/r/A*(1-EP))*r*r/2+tS*y/C;    }    else if(fabs(x_mult(a,b,c))>=r*C||cross(b,c,a)<=0||cross(a,c,b)<=0){        if(cross(a,b,c)<0)            if(x_mult(a,b,c)<0)                return (-acos(-1.0)-asin(x_mult(a,b,c)/A/B*(1-EP)))*r*r/2;            else return (acos(-1.0)-asin(x_mult(a,b,c)/A/B*(1-EP)))*r*r/2;        else return asin(x_mult(a,b,c)/A/B*(1-EP))*r*r/2;    }    else{        x=(cross(a,c,b)+sqrt(r*r*C*C-x_mult(a,c,b)*x_mult(a,c,b)))/C;        y=(cross(b,c,a)+sqrt(r*r*C*C-x_mult(b,c,a)*x_mult(b,c,a)))/C;        tS=x_mult(a,b,c)/2;        return (asin(tS*(1-x/C)*2/r/B*(1-EP))+asin(tS*(1-y/C)*2/r/A*(1-EP)))*r*r/2+tS*((y+x)/C-1);    }}double solve(Point p[], int n, Point cir, double r){    double area=0;    for(int i=0;i<n;i++){        area+=cal_area(p[i], p[(i+1)%n], cir, r);    }    return area;}double pi = acos(-1.0);int main(){    double v0, v1, D, T;int cas = 1;    while(cin >> v0 >> v1 >> D >> T)    {        printf("Case %d: ", cas ++);        Point cir = Point(-D, 0);        double r = v0 * T;        if(v0 * T <= D)        {            printf("%.7f\n", pi * r * r);            continue;        }        double lft = D / v0, rgt = T;        for(int i = 0; i < 100; i ++)        {            double mid = (lft + rgt) / 2;            double midmid = (mid + rgt) / 2;            double _mid = sqrt((v0 * mid) * (v0 * mid) - D * D) + v1 * (T - mid);            double _midmid = sqrt((v0 * midmid) * (v0 * midmid) - D * D) + v1 * (T - midmid);            if(_mid > _midmid) rgt = midmid;            else lft = mid;        }        double t = (lft + rgt) / 2;        double y = sqrt((v0 * t) * (v0 * t) - D * D) + v1 * (T - t);        double x1 = T / t * D, y1 = sqrt(v0 * T * v0 * T - x1 * x1);        x1 -= D;        Point p[8];        p[0] = Point(0, y);        p[1] = Point(x1, y1);        p[2] = Point(x1, -y1);        p[3] = Point(0, -y);        p[4] = Point(-x1, -y1);        p[5] = Point(-x1, y1);        p[6] = p[0];        double ans = solve(p, 6, cir, r);        ans = pi * r * r + x1 * y1 * 4 + x1 * (y - y1) * 2 - fabs(ans);        printf("%.7f\n", ans);    }}