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 = 400000+10; 
const double PI = acos(-1.0);
const double eps = 1e-12;

int dcmp(double x) {
    if(fabs(x)<eps) return 0; else return x<0? -1:1;
}

struct Pt {
    double x,y;
    Pt(double x=0,double y=0) :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)==0 && dcmp(a.y-b.y)==0;
}
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=0;
    vector<Pt> ch(n+1);
    for(int i=0;i<n;i++) {
        while(m>1 && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
        ch[m++]=p[i];
    }
    int k=m;
    for(int i=n-2;i>=0;i--) {
        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--;
    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-=2*r , b-=2*r; a/=2 , b/=2;
    double x,y,ang,ans=0;
    FOR(i,1,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,0,n-2)
        ans+=dist(ch[i],ch[i+1]);
    ans+=dist(ch[n-1],ch[0]);
    ans+=2*PI*r;
    printf("%.2lf",ans);
    return 0;
}