Bzoj2829 信用卡凸包

Time Limit: 10 Sec  Memory Limit: 128 MBSec  Special Judge
Submit: 333  Solved: 155

Description

Input

Output

Sample Input

2
6.0 2.0 0.0
0.0 0.0 0.0
2.0 -2.0 1.5707963268

Sample Output

21.66

HINT

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)

Source

几何 凸包

把矩形边长减去圆的直径，得到有效边长。计算出每个矩形的四个点，找到凸包，再加上一个圆的周长就是答案。

迷之WA，注释掉的旋转方法不知为何不对（小样例居然都过了）

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const double Pi=acos(-1.0);
const int mxn=100010;
struct point{
    double x,y;
    point operator - (point rhs){return (point){x-rhs.x,y-rhs.y};}
    point operator + (point rhs){return (point){x+rhs.x,y+rhs.y};}
}p[mxn];
 
double cross(point a,point b){return a.x*b.y-a.y*b.x;}
double dist(point a,point b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}
int cmp(point a,point b){
    double tmp=cross(a-p[1],b-p[1]);
    return (tmp<0 || (tmp==0 && dist(a,p[0])>dist(b,p[0])));
}
int cmp2(point a,point b,point p){
    double tmp=cross(a-p,b-p);
    return (tmp<0 || (tmp==0 && dist(a,p)>=dist(b,p)));
}
point rotate(point a,double angle){
    return (point){a.x*cos(angle)-a.y*sin(angle),a.x*sin(angle)+a.y*cos(angle)};
}
point move(point a,double len,double an){
    return (point){a.x+len*cos(an),a.y+len*sin(an)};
}
int n;
double a,b,r;
double x,y,th;
int st[mxn],top;
int main(){
    int i,j;
    scanf("%d",&n);
    scanf("%lf%lf%lf",&a,&b,&r);
    a-=2*r;b-=2*r;
    top=0;
    for(i=1;i<=n;i++){
        scanf("%lf%lf%lf",&x,&y,&th);
        for(j=1;j<=4;j++)
        {
            point tmp=move((point){x,y},b/2,th+Pi*(j-1)/2);
            p[++top]=move(tmp,a/2,th+Pi*j/2);
            swap(a,b);
        }        
//        point tmp;
/*
        tmp.x=a/2;tmp.y=b/2;p[++top]=(point){x,y}+rotate(tmp,th);
        tmp.x=-a/2;tmp.y=b/2;p[++top]=(point){x,y}+rotate(tmp,th);
        tmp.x=a/2;tmp.y=-b/2;p[++top]=(point){x,y}+rotate(tmp,th);
        tmp.x=-a/2;tmp.y=-b/2;p[++top]=(point){x,y}+rotate(tmp,th);
*/
    }
    n=top;top=0;
    int s=1;
    for(i=1;i<=n;i++){
        if(p[i].x<p[s].x || (p[i].x==p[s].x && p[i].y<p[s].y))s=i;
    }
    swap(p[s],p[1]);
    sort(p+2,p+n+1,cmp);
    p[n+1]=p[1];
    n++;
    for(i=1;i<=n;i++){
        while(top>1 && cmp2(p[i],p[st[top]],p[st[top-1]]))top--;
        st[++top]=i;
    }
    double ans=0;
    ans+=Pi*2*r;
    for(i=1;i<top;i++){
        ans+=sqrt(dist(p[st[i]],p[st[i+1]]));
    }
    printf("%.2lf
",ans);
    return 0;
}
/*
5
12.0 6.0 2.0
0.0 0.0 0.0
1 1 1
2 2 1.5707963268
1 7 0
2.0 -2.0 1.5707963268
*/