简单几何(向量旋转+凸包+多边形面积) UVA 10652 Board Wrapping

 

题目传送门

题意：告诉若干个矩形的信息，问他们在凸多边形中所占的面积比例

分析：训练指南P272，矩形面积长*宽，只要计算出所有的点，用凸包后再求多边形面积。已知矩形的中心，向量在原点参考点再旋转，角度要转换成弧度制。

 

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/10 星期二 10:34:43
* File Name     :UVA_10652.cpp
 ************************************************/

#include <bits/stdc++.h>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x)  {
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
struct Point    {
    double x, y;
    Point () {}
    Point (double x, double y) : x (x), y (y) {}
    Point operator - (const Point &r) const {       //向量减法
        return Point (x - r.x, y - r.y);
    }
    Point operator * (double p) const {       //向量乘以标量
        return Point (x * p, y * p);
    }
    Point operator / (double p) const {       //向量除以标量
        return Point (x / p, y / p);
    }
    Point operator + (const Point &r) const {
        return Point (x + r.x, y + r.y);
    }
    bool operator < (const Point &r) const {
        return x < r.x || (x == r.x && y < r.y);
    }
    bool operator == (const Point &r) const {
        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
    }
};
typedef Point Vector;
double dot(Vector A, Vector B)  {       //向量点积
    return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B)    {       //向量叉积
    return A.x * B.y - A.y * B.x;
}
Vector rotate(Vector A, double rad)   {
    return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
double area_poly(vector<Point> ps)  {
    double ret = 0;
    for (int i=1; i<ps.size ()-1; ++i)  {
        ret += fabs (cross (ps[i] - ps[0], ps[i+1] - ps[0])) / 2;
    }
    return ret;
}
vector<Point> convex_hull(vector<Point> ps) {
    sort (ps.begin (), ps.end ());
    ps.erase (unique (ps.begin (), ps.end ()), ps.end ());
    int n = ps.size (), k = 0;
    vector<Point> qs (n * 2);
    for (int i=0; i<n; ++i) {
        while (k > 1 && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-2]) <= 0)    k--;
        qs[k++] = ps[i];
    }
    for (int t=k, i=n-2; i>=0; --i) {
        while (k > t && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-2]) <= 0)    k--;
        qs[k++] = ps[i];
    }
    qs.resize (k - 1);
    return qs;
}

int main(void)    {
    int T;  scanf ("%d", &T);
    while (T--) {
        int n;  scanf ("%d", &n);
        vector<Point> ps;
        double area1 = 0;
        double x, y, w, h, r;
        for (int i=0; i<n; ++i) {
            scanf ("%lf%lf%lf%lf%lf", &x, &y, &w, &h, &r);
            Point a = Point (x, y);
            area1 += w * h;
            r = -r / 180 * PI;
            ps.push_back (a + rotate (Vector (-w/2, -h/2), r));
            ps.push_back (a + rotate (Vector (w/2, -h/2), r));
            ps.push_back (a + rotate (Vector (w/2, h/2), r));
            ps.push_back (a + rotate (Vector (-w/2, h/2), r));
        }
        vector<Point> qs = convex_hull (ps);
        printf ("%.1f %%\n", 100 * area1 / area_poly (qs));
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";

    return 0;
}