[SCOI2015]小凸想跑步

题目描述

小凸晚上喜欢到操场跑步，今天他跑完两圈之后，他玩起了这样一个游戏。

操场是个凸

现在小凸想知道他一次站位正确的概率是多少。

题解

我们其实是要找到一个p点，使得pp0*pp1<=ppi*ppi+1.

然后我们把上面的式子展开，然后化简，这个不难就是挺麻烦的。

最后得到了Ax+By+C<=0的形式，然后可以用(-1e9,y1)(1e9,y2)这条直线来描述这个限制，再加上凸多边形的限制，跑个半平面交就好了。

注意：要特判A或B=0的情况。

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#define N 200009
#define double long double
#define eq(x,y) (fabs((x)-(y))<eps)
using namespace std;
const double eps=1e-15;
int n,top,tot;
double x[N],y[N],s,S;
inline int rd(){
    int x=0;char c=getchar();bool f=0;
    while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
struct point{
    double x,y;
    point(double xx=0,double yy=0){x=xx;y=yy;}
    inline point operator +(const point &b)const{return point{x+b.x,y+b.y};} 
    inline point operator -(const point &b)const{return point{x-b.x,y-b.y};}
    inline double operator *(const point &b)const{return x*b.y-y*b.x;}
    inline point operator *(const double &b)const{return point{x*b,y*b};} 
}p[N];
struct line{
    point x,y;double ang;
    line(double x1=0,double x2=0,double x3=0,double x4=0){x.x=x1;x.y=x2;y.x=x3;y.y=x4;}
     bool operator <(const line &b)const{
        if(fabs(ang-b.ang)<eps)return (y-x)*(b.x-x)<eps; 
        else return ang<b.ang;
    }
}a[N],l[N],q[N];
inline bool left(point a,line b){return (a-b.x)*(b.y-b.x)>-eps;}
inline point jiao(line a,line b){return b.x+(b.y-b.x)*(((b.x-a.x)*(a.y-a.x))/((a.y-a.x)*(b.y-b.x)));}
int main(){
    n=rd();
    for(int i=0;i<n;++i){
        x[i]=rd(),y[i]=rd();
        if(i)l[++top]=line(x[i-1],y[i-1],x[i],y[i]);
    }
    l[++top]=line(x[n-1],y[n-1],x[0],y[0]);
    for(int i=2;i<n;++i)S+=(point(x[i],y[i])-point(x[0],y[0]))*(point{x[i-1],y[i-1]}-point{x[0],y[0]})/2;S=fabs(S);
    x[n]=x[0];y[n]=y[0];
    for(int i=1;i<n;++i){
        double a=-y[1]+y[0]+y[i+1]-y[i],b=-x[0]+x[1]+x[i]-x[i+1],c=x[0]*y[1]-x[1]*y[0]-x[i]*y[i+1]+x[i+1]*y[i];
        if(fabs(b)<eps){
            if(fabs(a)<eps){puts("0.0000");return 0;}
            double xf=-c/a,xs=xf,yf=0,ys=1e15;
            if(a<0)l[++top]=line(xs,ys,xf,yf);
            else l[++top]=line(xf,yf,xs,ys);
        }
        else{
           double xf=-1e11,xs=1e11,yf=(-c-a*xf)/b,ys=(-c-a*xs)/b;
           if(b>=0){l[++top]=line(xs,ys,xf,yf);
           }else l[++top]=line(xf,yf,xs,ys);
        }
    }
    for(int i=1;i<=top;++i)l[i].ang=atan2(l[i].y.y-l[i].x.y,l[i].y.x-l[i].x.x);
    sort(l+1,l+top+1);
    for(int i=1;i<=top;++i)if(i==1||fabs(l[i].ang-l[i-1].ang)>eps)a[++tot]=l[i];
    int h=1,t=2;q[1]=a[1];q[2]=a[2];p[1]=jiao(a[1],a[2]);
    for(int i=3;i<=tot;++i){
        while(h<t&&left(p[t-1],a[i]))t--;
        while(h<t&&left(p[h],a[i]))h++;
        q[++t]=a[i];p[t-1]=jiao(q[t-1],q[t]);
    }
    while(h<t&&left(p[h],q[t]))h++;
    while(h<t&&left(p[t-1],q[h]))t--;
    p[t]=jiao(q[t],q[h]);
    for(int i=h+2;i<=t;++i)s+=(p[i]-p[h])*(p[i-1]-p[h])/2;s=fabs(s);
    printf("%.4LF",s/S);
    return 0;
}

posted @ 2019-02-24 14:59 comld 阅读(294) 评论(0) 编辑收藏举报

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[SCOI2015]小凸想跑步

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