BZOJ 1043 圆弧覆盖

为按照极角，转化为线段覆盖问题。

这题有个特别混蛋的trick，就是后掉落的把先掉落的完全覆盖了，我就没考虑这个，无限wa啊！！

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

#define N 1111
#define PI 3.14159265358979323846
#define EPS 1e-7

using namespace std;

struct PO
{
    double x,y,r;
    void prt() {printf("%lf     %lf\n",x,y);}
}p[N];

struct V
{
    int fg; double f;
};
vector<V> v[N];

int n;
double sum,ans;
bool vis[N];

inline int dc(double x)
{
    if(x>EPS) return 1;
    else if(x<-EPS) return -1;
    return 0;
}

inline bool cmp(const V &a,const V &b)
{
    if(a.f==b.f) return a.fg>b.fg;
    return a.f<b.f;
}

inline PO operator -(PO a,PO b)
{
    a.x-=b.x; a.y-=b.y;
    return a;
}

inline PO operator +(PO a,PO b)
{
    a.x+=b.x; a.y+=b.y;
    return a;
}

inline PO operator *(PO a,double k)
{
    a.x*=k; a.y*=k;
    return a;
}

inline PO operator /(PO a,double k)
{
    a.x/=k; a.y/=k;
    return a;
}

inline double getlen(const PO &a)
{
    return sqrt(a.x*a.x+a.y*a.y);
}

inline double getdis(const PO &a,const PO &b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline PO rotate(const PO &a,double sss,double ccc)
{
    PO rt;
    rt.x=a.x*ccc-a.y*sss;
    rt.y=a.x*sss+a.y*ccc;
    return rt;
}

inline void read()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lf%lf%lf",&p[i].r,&p[i].x,&p[i].y);
}

inline void getcpoint(const PO &a,const PO &b,PO &s,PO &t)//求圆的交点，a的逆时针方向的弧上的点 
{
    PO ab=b-a;
    double d=getlen(ab);
    double ccc=(b.r*b.r-d*d-a.r*a.r)/(-2.0*a.r*d);
    double sss=sqrt(1.0-ccc*ccc);
    ab=ab/d*a.r;
    s=a+rotate(ab,-sss,ccc);
    t=a+rotate(ab,sss,ccc);
}

inline void go()
{
    PO a,b; V c;
    double f1,f2;
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
        {
            double dis=getdis(p[i],p[j]);
            if(dis+1e-10>p[i].r+p[j].r || p[j].r+1e-10<p[i].r&&dis-1e-10<p[i].r-p[j].r) continue;
            if(p[i].r<p[j].r+1e-10&&dis-1e-10<p[j].r-p[i].r) {vis[i]=true; break;}
            getcpoint(p[i],p[j],a,b);
            f1=atan2(a.y-p[i].y,a.x-p[i].x);
            f2=atan2(b.y-p[i].y,b.x-p[i].x);
            if(f1>f2)
            {
                c.f=f1; c.fg=1; v[i].push_back(c);
                c.f=PI; c.fg=-1; v[i].push_back(c);
                c.f=-PI; c.fg=1; v[i].push_back(c);
                c.f=f2; c.fg=-1; v[i].push_back(c);
            }
            else
            {
                c.f=f1; c.fg=1; v[i].push_back(c);
                c.f=f2; c.fg=-1; v[i].push_back(c);
            }
        }
    for(int i=1;i<=n;i++)
    {
        if(vis[i]) continue;
        sum+=2*PI*p[i].r;
        sort(v[i].begin(),v[i].end(),cmp);
        int tmp=0; double last;
        for(int j=0;j<v[i].size();j++)
        {
            tmp+=v[i][j].fg;
            if(tmp==1&&v[i][j].fg==1) last=v[i][j].f;
            else if(tmp==0&&v[i][j].fg==-1) ans+=(v[i][j].f-last)*p[i].r;
        }
    }
    printf("%.3lf\n",sum-ans);
}

int main()
{
    read(),go();
    return 0;
}