Description
给定很多块木板,用面积尽量小的多边形将其包起来。
Solution
将木板的四个点全部加入点集,求凸包即可。
Source
/****************************
* Au: Hany01
* Prob: [UVa 10652] Board Wrapping
* Date: Feb 12th, 2018
* Email: hany01@foxmail.com
****************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define rep(i , j) for (int i = 0 , i##_end_ = j; i < i##_end_ ; ++ i)
#define For(i , j , k) for (int i = (j) , i##_end_ = (k) ; i <= i##_end_ ; ++ i)
#define Fordown(i , j , k) for (int i = (j) , i##_end_ = (k) ; i >= i##_end_ ; -- i)
#define Set(a , b) memset(a , b , sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define y1 wozenmezhemecaia
#ifdef hany01
#define debug(...) fprintf(stderr , __VA_ARGS__)
#else
#define debug(...)
#endif
inline void File() {
#ifdef hany01
freopen("uva10652.in" , "r" , stdin);
freopen("uva10652.out" , "w" , stdout);
#endif
}
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }
inline int read() {
register char c_; register int _ , __;
for (_ = 0 , __ = 1 , c_ = getchar() ; !isdigit(c_) ; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; isdigit(c_) ; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}
const double Pi = acos(-1.), eps = 1e-9;
const int maxn = 2405;
int n, m;
inline int dcmp(double x) { if (fabs(x) < eps) return 0; return x < 0 ? -x : x; }
struct Point
{
double x, y;
Point(double x = 0, double y = 0): x(x), y(y) {}
}p[maxn], q[maxn];
typedef Point Vector;
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
bool operator < (const Vector& A, const Vector& B) { return A.x < B.x || (A.x == B.x && A.y - B.y < 0); }
double Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; }
Vector Rotate(Vector V, double rad) { return Vector(V.x * cos(rad) - V.y * sin(rad), V.x * sin(rad) + V.y * cos(rad)); }
int ConvexHull(Point* p, int n, Point* q) {
register int m = 0, tmp;
sort(p + 1, p + 1 + n);
For(i, 1, n) {
while (m > 1 && Cross(q[m] - q[m - 1], p[i] - q[m - 1]) <= 0) -- m;
q[++ m] = p[i];
}
tmp = m;
Fordown(i, n - 1, 1) {
while (m > tmp && Cross(q[m] - q[m - 1], p[i] - q[m - 1]) <= 0) -- m;
q[++ m] = p[i];
}
if (n > 1) -- m;
return m;
}
double S(Point* p, int n) {
double Ans = 0;
For(i, 2, n - 1) Ans += Cross(p[i] - p[1], p[i + 1] - p[1]);
return fabs(Ans / 2);
}
int main()
{
File();
for (register int T = read(); T --; ) {
register double w, h, j, Ans1 = 0, Ans2; register Point O; register int cnt = 0;
n = read();
For(i, 1, n) {
scanf("%lf%lf%lf%lf%lf", &O.x, &O.y, &w, &h, &j), j = -j * Pi / 180;
p[++ cnt] = O + Rotate(Vector(w / 2, h / 2), j), p[++ cnt] = O + Rotate(Vector(-w / 2, h / 2), j),
p[++ cnt] = O + Rotate(Vector(w / 2, -h / 2), j), p[++ cnt] = O + Rotate(Vector(-w / 2, -h / 2), j);
Ans1 += w * h;
}
n = cnt, m = ConvexHull(p, n, q);
Ans2 = S(q, m);
printf("%.1lf %c\n", Ans1 * 100 / Ans2, '%');
}
return 0;
}