洛谷-P3829-[SHOI2012]信用卡凸包（凸包，注意精度！！！！！！）

题目链接：https://www.luogu.org/problemnew/show/P3829

题目大意：中文题，易理解。

思路：Fuck！！精度卡了我一页的WA...（还是太菜了）经过简单的草图，如下图所示：随便举一个例子，然后发现，周长就是里面的小的多边形的周长+圆的周长。至于具体的证明，这不是显而易见的么？？然后就是求出每个矩形的四个点，然后求出凸包，就行了，简单的板子题，半个小时搞定！

好了好了！！板子确实好写，主要是精度！！四个点的坐标。。一开始我用的：

            Ang1=atan2(b,a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));
            Ang1=atan2(b,-a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));
            Ang1=atan2(-b,a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));
            Ang1=atan2(-b,-a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));

结果WA了5，6，10三个测试点。我还以为是我板子的问题，调了一节课（毛概一整节课都没调出来啊）。回寝室之后继续调，结果还是没有跳出来，我又写了一个Jaris法，还是5，6，10测试点。。这时我还以为是板子的问题（呜呜呜。。。

结果早上试着用其他方法得到四个点，发现就过了。。。万恶的精度啊！！不过也是，三角函数用了个遍，精度肯定损失到爆炸...

ACCode：

// luogu-judger-enable-o2
//#pragma comment(linker, "/STACK:1024000000,1024000000")
 
#include<stdio.h>
#include<string.h>
#include<math.h>
 
#include<map>
#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<string>
#include<fstream>
#include<iostream>
#include<algorithm>
using namespace std;
 
#define ll long long
#define Pair pair<int,int>
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// 水印
//std::ios::sync_with_stdio(false);
//  register
const int MAXN=1e4+10;
const int INF32=0x3f3f3f3f;
const ll INF64=0x3f3f3f3f3f3f3f3f;
const int MOD=1e9+7;
const double EPS=1.0e-8;
const double PI=acos(-1.0);

struct Point{
    double x,y;
    Point(double _x=0,double _y=0){
        x=_x;y=_y;
    }
    friend Point operator + (const Point &a,const Point &b){
        return Point(a.x+b.x,a.y+b.y);
    }
    friend Point operator - (const Point &a,const Point &b){
        return Point (a.x-b.x,a.y-b.y);
    }
    friend double operator ^ (const Point &a,const Point &b){
        return a.x*b.y-a.y*b.x;
    }
    friend bool operator == (const Point &a,const Point &b){
        return fabs(a.x-b.x)<EPS&&fabs(a.y-b.y)<EPS;
    }
};
struct V{
    Point start,end;
    V(Point _start=Point(0,0),Point _end=Point(0,0)){
        start=_start;end=_end;
    }
};
Point Dots[MAXN<<2];
Point Stk[MAXN<<2];int Top;
double A,B,R;
int N,Ecnt;

double Distance(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double Prellel(double key){
    return fabs(key)<EPS?0:key;
}
int Cmp(Point a,Point b){
    double res=Prellel((a-Dots[1])^(b-Dots[1]));
    if(res>0) return 1;
    if(res==0&&Distance(a,Dots[1])<Distance(b,Dots[1])) return 1;
    return 0;
}
int Cmp2(Point a,Point b){
	return (a.y<b.y||(a.y==b.y&&a.x<b.x));
}
int Judge(int k,int i){
    if(Dots[i].y<Dots[k].y||(Dots[i].y==Dots[k].y&&Dots[i].x<Dots[k].x)) return 1;
    return 0;
}
void Graham(){
	int k=1;
    for(int i=2;i<=Ecnt;++i){
//		cout<<"DOts: "<<Dots[i].x<<" "<<Dots[i].y<<endl;
    	if(Judge(k,i)) k=i;
    }swap(Dots[1],Dots[k]);
    sort(Dots+2,Dots+1+Ecnt,Cmp);
    Top=0;Stk[++Top]=Dots[1];
    for(int i=2;i<=Ecnt;++i){
    	while(Top>=3&&((Dots[i]-Stk[Top-1])^(Stk[Top]-Stk[Top-1]))>=0) --Top;
//        while(Top>=3&&((Stk[Top]-Stk[Top-1])^(Dots[i]-Stk[Top-1]))<EPS) --Top;
        Stk[++Top]=Dots[i];
    }//Stk[Top+1]=Stk[1];
}
void Jarvis(){
	int len;Top=0;
	sort(Dots+1,Dots+1+N,Cmp2);
	Stk[++Top]=Dots[1];Stk[++Top]=Dots[2];
	for(int i=3;i<=N;++i){
		while(Top>=2&&((Dots[i]-Stk[Top-1])^(Stk[Top-1]-Stk[Top]))>=0) --Top;
		Stk[++Top]=Dots[i];
	}int Bot=Top;
	for(int i=N-1;i>0;--i){
		while(Top>Bot&&((Dots[i]-Stk[Top-1])^(Stk[Top-1]-Stk[Top]))>=0) --Top;
		Stk[++Top]=Dots[i];
	}Stk[Top+1]=Stk[1];
}
Point rotate(Point a,double Ang){
    double c=cos(Ang),s=sin(Ang);
    return Point(a.x*c-a.y*s,a.x*s+a.y*c);
}
int main(){
    while(~scanf("%d",&N)){
        scanf("%lf%lf%lf",&A,&B,&R);
//		double a=A/2.0,b=B/2.0,r=R;
//        double a=A/2.0-R,b=B/2.0-R;
		double a=A-R*2.0,b=B-R*2.0;
//		cout<<a<<" "<<b<<endl;
        double C=sqrt(a*a+b*b);
        double x,y,Ang;Ecnt=0;
        double Ang1;
//		cout<<"C is : "<<C<<endl;
        for(int i=1;i<=N;++i){
            scanf("%lf%lf%lf",&x,&y,&Ang);
            
            Point k(x,y),d,v;double rad=Ang;
			d=Point(x+b/2,y-a/2),v=rotate(k-d,rad),Dots[++Ecnt]=k+v;
			d=Point(x-b/2,y-a/2),v=rotate(k-d,rad),Dots[++Ecnt]=k+v;
			d=Point(x+b/2,y+a/2),v=rotate(k-d,rad),Dots[++Ecnt]=k+v;
			d=Point(x-b/2,y+a/2),v=rotate(k-d,rad),Dots[++Ecnt]=k+v;
//			Dots[++Ecnt]=rotate(Point(a-r,b-r),Ang)+Point(x,y);
//			Dots[++Ecnt]=rotate(Point(r-a,b-r),Ang)+Point(x,y);
//			Dots[++Ecnt]=rotate(Point(r-a,r-b),Ang)+Point(x,y);
//			Dots[++Ecnt]=rotate(Point(a-r,r-b),Ang)+Point(x,y);
//            Ang1=atan2(b,a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));
//            Ang1=atan2(b,-a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));
//            Ang1=atan2(-b,a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));
//            Ang1=atan2(-b,-a);Dots[++Ecnt]=Point(x+C*cos(Ang+Ang1),y+C*sin(Ang+Ang1));
        }Graham();//Jarvis();
		double ans=2.0*R*PI;
        for(int i=2;i<=Top;++i){
            ans+=Distance(Stk[i],Stk[i-1]);
//			cout<<"the Stk is : "<<ans<<" "<<Stk[i].x<<" "<<Stk[i].y<<endl;
        }ans+=Distance(Stk[1],Stk[Top]);
		printf("%.2lf\n",ans);
    }
}

/*


*/

标签：Dots,Ang,洛谷,Top,凸包,Stk,Ang1,include,SHOI2012
来源： https://blog.csdn.net/qq_40482358/article/details/88375333