关于三分的暴力技巧（暴力好）

Posted 2020-11-24 starseven

关于三分的暴力技巧

以下为暴力专场（前方高能）

现在以hdu4454为例

1.这道题的一个总结点是暴力对于精度不超过1e-2的东西，我们可以直接暴力求解

2.几何：对于一个圆（整个几何）都需要掌握三角函数的运用，这里有一篇非常不错的文章可以学习

3.圆到矩形的最短距离

关于圆到矩形的最短距离，主要是关于是到

矩形上垂直一点到圆最小(这时x和y都被“包围”)；

矩形的四个顶点

具体看代码实现

现在附上超级牛逼的暴力代码（感谢qt神犇）

#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdlib>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-') f=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        x=(x<<3)+(x<<1)+ch-'0';
        ch=getchar();
    }  
    return x*f;
}
inline void out()
{
    fclose(stdin);
    fclose(stdout);
}
const double pi=acos(-1.0);
struct Point{
    double x,y,r;
}C,P,st;
double xx,yy,x2,y2,miny,maxy,minx,maxx;
inline double dist(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
inline double Rdist(Point a)
{
    double x=0.0,y=0.0;
    if(a.x<minx) x=minx-a.x;
    else if(a.x>maxx) x=a.x-maxx;
    if(a.y<miny) y=miny-a.y;
    else if(a.y>maxy) y=a.y-maxy;
    return sqrt(x*x+y*y);
}
int main(void)
{
    while(1)
    {
        scanf("%lf%lf",&st.x,&st.y);
        if(st.x==0&&st.y==0) break;                     
        double ans=1e10;
        scanf("%lf%lf%lf",&C.x,&C.y,&C.r);
        scanf("%lf%lf%lf%lf",&xx,&yy,&x2,&y2);
        minx=min(xx,x2),maxx=max(xx,x2);
        miny=min(yy,y2),maxy=max(yy,y2);
        for(double rad=0;rad<=360;rad+=0.005)
        {
            Point P;                                                               
            P.x=C.x+C.r*cos(rad/180*pi);
            P.y=C.y+C.r*sin(rad/180*pi);
            ans=min(ans,dist(P,st)+Rdist(P));
        }
        printf("%.2f
",ans);
    }
    out();
    return 0;
}

还有正解（所谓的正解都是忽悠老实人的）

#include<algorithm>
#include<iostream>
#include<fstream>
#include<sstream>
#include<cstring>
#include<cstdlib>
#include<string>
#include<vector>
#include<cstdio>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define LL long long
#define PB push_back
#define eps 1e-10
#define debug puts("**debug**");
using namespace std;
const double PI = acos(-1);
 
struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x), y(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); }
bool operator < (const Point& a, const Point& b)
{
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
int dcmp(double x)
{
    if(fabs(x) < eps) return 0; return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b)
{
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}
 
double Dot(Vector a, Vector b) { return a.x*b.x + a.y*b.y; }
double Length(Vector a) { return sqrt(Dot(a, a)); }
double Cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x; }
double DistanceToSegment(Point p, Point a, Point b)
{
    if(a == b) return Length(p-a);
    Vector v1 = b-a, v2 = p-a, v3 = p-b;
    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
    else return fabs(Cross(v1, v2)) / Length(v1);
}
struct Circle
{
    Point c;
    double r;
    Circle(){}
    Circle(Point c, double r):c(c), r(r){}
    Point point(double a) //根据圆心角求点坐标
    {
        return Point(c.x+cos(a)*r, c.y+sin(a)*r);
    }
}o;
 
Point p, p1, p2, p3, p4, s;
double a, b, c, d;
 
 
double Calc(double x)
{
    p = o.point(x);
    double d1 = DistanceToSegment(p, p1, p2),
    d2 = DistanceToSegment(p, p2, p3),
    d3 = DistanceToSegment(p, p3, p4),
    d4 = DistanceToSegment(p, p4, p1);
    //点p到矩形最近距离加上s到p距离
    return min(min(d1, d2), min(d3, d4)) + Length(s-p);
}
 
double solve()
{
    double L, R, m, mm, mv, mmv;
    L = 0; R = PI;
    while (L + eps < R)
    {
        m = (L + R) / 2;
        mm = (m + R) / 2;
        mv = Calc(m);
        mmv = Calc(mm);
        if (mv <= mmv) R = mm; //三分法求最大值时改为mv>=mmv
        else L = m;
    }
    double ret = Calc(L);
    L = PI; R = 2*PI;
    while (L + eps < R)
    {
        m = (L + R) / 2;
        mm = (m + R) / 2;
        mv = Calc(m);
        mmv = Calc(mm);
        if (mv <= mmv) R = mm; 
        else L = m;
    }
    return min(ret, Calc(L));
}
 
int main()
{
    while(scanf("%lf%lf", &s.x, &s.y))
    {
        if(s.x == 0 && s.y == 0) break;
        scanf("%lf%lf%lf", &o.c.x, &o.c.y, &o.r);
        scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
        //确定矩形四个点坐标，左上点开始 逆时针
        double maxx, maxy, minx, miny;
        maxx = max(a, c); maxy = max(b, d);
        minx = min(a, c); miny = min(b, d);
        p1 = Point(minx, maxy);
        p2 = Point(minx, miny);
        p3 = Point(maxx, miny);
        p4 = Point(maxx, maxy);
        double ans = solve();
        printf("%.2f
", ans);
    }
    return 0;
}