[hdu4667]Building Fence 计算几何 瞎瘠薄搞

大致题意：

　　给出n个圆和m个三角形，求最小的的，能将所有图形覆盖的图形的周长。

　　正解为求所有三角形顶点与圆的切点以及圆和圆的切点构造凸包，再求路径。

  　 因为要求结果误差<=1e-3 所以 我们可以在每个圆上，将圆分为1000(或更多)等分，得到每个圆上有1000个点，把这些点和三角形的顶点构造凸包，为减小误差，圆上两点的距离依然为圆弧的长度。这样累加边即可得答案。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<cstdlib>
#include<cmath>
#include<list>
using namespace std;
#define MAXN 100100
#define eps 1e-9
#define For(i,a,b) for(int i=a;i<=b;i++) 
#define Fore(i,a,b) for(int i=a;i>=b;i--) 
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mkp make_pair
#define pb push_back
#define cr clear()
#define sz size()
#define met(a,b) memset(a,b,sizeof(a))
#define iossy ios::sync_with_stdio(false)
#define fre freopen
#define pi acos(-1.0)
#define inf 1e6+7
#define Vector Point
const int Mod=1e9+7;
typedef unsigned long long ull;
typedef long long ll;
int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x<0?-1:1;
}
struct Point{
    double x,y;
    int id;
    Point(double x=0,double y=0,int id=-1):x(x),y(y),id(id) {}
    bool operator < (const Point &a)const{
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
    Point operator - (const Point &a)const{
        return Point(x-a.x,y-a.y);
    }
    Point operator + (const Point &a)const{
        return Point(x+a.x,y+a.y);
    }
    Point operator * (const double &a)const{
        return Point(x*a,y*a);
    }
    Point operator / (const double &a)const{
        return Point(x/a,y/a);
    }
    void read(){
        scanf("%lf%lf",&x,&y);
    }
    void out(){
        cout<<"debug: "<<x<<" "<<y<<endl;
    }
    bool operator == (const Point &a)const{
        return dcmp(x-a.x)==0 && dcmp(y-a.y)==0;
    }
};
double Dot(Vector a,Vector b) {
    return a.x*b.x+a.y*b.y;
}
double dis(Vector a) {
    return sqrt(Dot(a,a));
}
double Cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
int ConvexHull(Point *p,int n,Point *ch){
    int m=0;
    For(i,0,n-1) {
        while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    int k=m;
    Fore(i,n-2,0){
        while(m>k && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    if(n>1) m--;
    return m;
}
void Swap(int &p1,int &p2){
    p1^=p2;
    p2^=p1;
    p1^=p2;
}
int n,m;
Point p[1000005];
Point ch[1000005];
Point cp[105];
double r[105];
void solve(){
    int rt=0;
    For(i,0,n-1) cp[i].read(),scanf("%lf",&r[i]);
    For(i,0,m-1) p[rt].id=n,p[rt++].read(),p[rt].id=n,p[rt++].read(),p[rt].id=n,p[rt++].read();
    For(i,0,n-1) {
        For(j,0,999){
            double ang=2*j*pi*1.0/1000;
            p[rt++]=Point(cp[i].x+r[i]*cos(ang),cp[i].y+r[i]*sin(ang),i);
        }
    }
    sort(p,p+rt);
    m=ConvexHull(p,rt,ch);
    double ans=0;
    For(i,0,m-1) {
        if(ch[i].id==ch[(i+1)%m].id && ch[i].id!=n) {
            ans+=pi*2*r[ch[i].id]/1000;
        }
        else ans+=dis(ch[i]-ch[(i+1)%m]);
    }
    printf("%.5lf
",ans);
}
int main(){
//    fre("in.txt","r",stdin);
    int t=0;
    while(~scanf("%d%d",&n,&m)) solve();
    return 0;
}