bzoj 2829 信用卡凸包（凸包）

2829: 信用卡凸包

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 1342  Solved: 577
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Description

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Input

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Output

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Sample Input

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2
6.0 2.0 0.0
0.0 0.0 0.0
2.0 -2.0 1.5707963268

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Sample Output

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3.333

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HINT

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本样例中的2张信用卡的轮廓在上图中用实线标出，如果视1.5707963268为Pi/2(pi为圆周率),则其凸包的周长为16+4*sqrt(2)

【思路】

凸包

将一张信用卡看作以四个圆心为顶点的矩形，求出凸包周长后再加一个圆周长。

【代码】

 #include<cmath>
 #include<cstdio>
 #include<vector>
 #include<cstring>
 #include<algorithm>
 #define FOR(a,b,c) for(int a=(b);a<=(c);a++)
 using namespace std;

 const int N = +;
 const double PI = acos(-1.0);
 const double eps = 1e-;

 int dcmp(double x) {
     if(fabs(x)<eps) return ; else return x<? -:;
 }

 struct Pt {
     double x,y;
     Pt(double x=,double y=) :x(x),y(y) {};
 };
 typedef Pt vec;

 vec operator - (Pt a,Pt b) { return vec(a.x-b.x,a.y-b.y); }
 vec operator + (vec a,vec b) { return vec(a.x+b.x,a.y+b.y); }
 bool operator == (Pt a,Pt b) {
     return dcmp(a.x-b.x)== && dcmp(a.y-b.y)==;
 }
 bool operator < (const Pt& a,const Pt& b) {
     return a.x<b.x || (a.x==b.x && a.y<b.y);
 }

 vec rotate(vec a,double x) {
     return vec(a.x*cos(x)-a.y*sin(x),a.x*sin(x)+a.y*cos(x));
 }
 double cross(vec a,vec b) { return a.x*b.y-a.y*b.x; }
 double dist(Pt a,Pt b) {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }

 vector<Pt> ConvexHull(vector<Pt> p) {
     sort(p.begin(),p.end());
     p.erase(unique(p.begin(),p.end()),p.end());
     int n=p.size() , m=;
     vector<Pt> ch(n+);
     for(int i=;i<n;i++) {
         while(m> && cross(ch[m-]-ch[m-],p[i]-ch[m-])<=) m--;
         ch[m++]=p[i];
     }
     int k=m;
     for(int i=n-;i>=;i--) {
         while(m>k && cross(ch[m-]-ch[m-],p[i]-ch[m-])<=) m--;
         ch[m++]=p[i];
     }
     if(n>) m--;
     ch.resize(m); return ch;
 }

 int n;
 double a,b,r;
 vector <Pt> p,ch;

 int main() {
     scanf("%d%lf%lf%lf",&n,&b,&a,&r);
     a-=*r , b-=*r; a/= , b/=;
     double x,y,ang,ans=;
     FOR(i,,n) {
         scanf("%lf%lf%lf",&x,&y,&ang);
         Pt o=Pt(x,y);
         p.push_back(o+rotate(vec(-a,-b),ang));
         p.push_back(o+rotate(vec(-a,b),ang));
         p.push_back(o+rotate(vec(a,-b),ang));
         p.push_back(o+rotate(vec(a,b),ang));
     }
     ch=ConvexHull(p);
     n=ch.size();
     FOR(i,,n-)
         ans+=dist(ch[i],ch[i+]);
     ans+=dist(ch[n-],ch[]);
     ans+=*PI*r;
     printf("%.2lf",ans);
     return ;
 }