UVA 10652 Board Wrapping(計算幾何基礎，求凸包)

題目連結：傳送門 

分析：

沒有什麼好說的就是求一個凸包就好了。可以當作模板、

程式碼如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const double eps = 1e-10;

//判斷符號，提高精度
int dcmp(double x){
    if(fabs(x)<eps) return 0;
    else return x < 0 ? -1 : 1;
}

struct Point{
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) { }
    bool operator < (const Point& a) const{
        if(a.x != x) return x < a.x;
        return y < a.y;
    }
    bool operator == (const Point B)const{
        return dcmp(x-B.x)==0&&dcmp(y-B.y)==0;
    }
};

typedef Point Vector;
Vector operator + (Vector A, Vector B){
    return Vector(A.x+B.x, A.y+B.y);
}
Vector operator - (Point A, Point B){
    return Vector(A.x-B.x, A.y-B.y);
}
Vector operator * (Vector A, double p){
    return Vector(A.x*p, A.y*p);
}
Vector operator / (Vector A, double p){
    return Vector(A.x/p, A.y/p);
}
double Dot(Vector A, Vector B){//向量相乘
    return A.x*B.x + A.y*B.y;  //a*b*cos(a,b)
}
double Length(Vector A){
    return sqrt(Dot(A, A));    //向量的長度
}
double Angle(Vector A, Vector B){
    return acos(Dot(A, B) / Length(A) / Length(B));    //向量的角度
}
double Cross(Vector A, Vector B){//叉積
    return A.x*B.y - A.y*B.x;
}

/**
向量(x,y) 繞起點逆時針旋轉a度。
x' = x*cosa - y*sina
y' = x*sina + y*cosa
**/
Vector Rotate(Vector A,double a){
    return Vector (A.x*cos(a)-A.y*sin(a),A.x*sin(a)+A.y*cos(a));
}

double trans(double ang){
    return ang/180*acos(-1.0);
}

//叉積,可以用來判斷方向和求面積
double cross(Point a,Point b,Point c){
    return (c.x-a.x)*(b.y-a.y) - (b.x-a.x)*(c.y-a.y);
}

//求多邊形的面積
double S(Point p[],int n)
{
    double ans = 0;
    p[n] = p[0];
    for(int i=1; i<n; i++)
        ans += cross(p[0],p[i],p[i+1]);
    if(ans < 0) ans = -ans;
    return ans / 2.0;
}

/**
求二維凸包Andrew演算法，將所有的點按x小到大(x相等，y小到大)排序
刪去重複的點，得到一個序列p1,p2...,然後把p1,p2放入凸包中，從p3
開始當新點再前進方向左邊時(可以用叉積判斷方向)繼續，否則，依次
刪除最近加入凸包的點，直到新點再左邊。
**/

int ConvexHull(Point *p,int n,Point *stack){
    sort(p,p+n);
    n=unique(p,p+n)-p;
    int m=0;
    for(int i=0; i<n; i++) {//如果不希望凸包的邊上有輸入的點則把兩個等號去掉
        while(m>1&&cross(stack[m-2],p[i],stack[m-1])<=0) m--;
        stack[m++]=p[i];
    }
    int k=m;
    for(int i=n-2; i>=0; i--){
        while(m>k&&cross(stack[m-2],p[i],stack[m-1])<=0)m--;
        stack[m++]=p[i];
    }
    if(n>1) m--;
    return m;
}

Vector p[2500];
Vector st[2500];

int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        double x,y,w,h,a;
        int num = 0;
        double tot = 0;
        for(int i=0; i<n; i++){
            scanf("%lf%lf%lf%lf%lf",&x,&y,&w,&h,&a);
            double ang = -trans(a);
            Point o = Point(x,y);
            p[num++] = o + Rotate(Point(-w/2,-h/2),ang);
            p[num++] = o + Rotate(Point(w/2,-h/2),ang);
            p[num++] = o + Rotate(Point(-w/2,h/2),ang);
            p[num++] = o + Rotate(Point(w/2,h/2),ang);
            tot+=w*h;
        }
        int m = ConvexHull(p,num,st);
        double area = S(st,m);
        printf("%.1lf %%\n",tot*100/area);
    }
    return 0;
}

/****

1
4
4 7.5 6 3 0
8 11.5 6 3 0
9.5 6 6 3 90
4.5 3 4.4721 2.2361 26.565

****/