conv cut(conv &a)
用此半平面切割凸包,返回新的凸包conv core(poly &a)
返回多边形a
的核(站在核中的点,可以看到多边形内的任意一点)conv conv_inter(vector<poly> &a)
返回 n 个多边形的交(多边形需要逆时针存储)
题目大意:
给 n 个凸多边形,求其面积交。
namespace Halfplane {
struct halfplane {
//ax+by+c<=0
db a, b, c;
halfplane(pt p, pt q) {
a = p.y - q.y;
b = q.x - p.x;
c = det(p, q);
}
halfplane(db aa, db bb, db cc) {
a = aa; b = bb; c = cc;
}
db calc(pt &t) {
return t.x * a + t.y * b + c;
}
pt intersect(pt &a, pt &b) {
pt res;
double t1 = calc(a), t2 = calc(b);
res.x = (t2 * a.x - t1 * b.x) / (t2 - t1);
res.y = (t2 * a.y - t1 * b.y) / (t2 - t1);
return res;
}
conv cut(conv &a) {
int n = (int)a.P.size();
conv res;
for (int i = 0; i < n; i++) {
if (calc(a.P[i]) < -eps) res.P.push_back(a.P[i]);
else {
int j = bef(i);
if (calc(a.P[j]) < -eps) res.P.push_back(intersect(a.P[j], a.P[i]));
j = nex(i);
if (calc(a.P[j]) < -eps) res.P.push_back(intersect(a.P[i], a.P[j]));
}
}
return res;
}
};
conv core(poly &a) {
conv res;
res.P.push_back(pt(-INF, -INF));
res.P.push_back(pt(INF, -INF));
res.P.push_back(pt(INF, INF));
res.P.push_back(pt(-INF, INF));
int n = (int)a.a.size();
for (int i = 0; i < n; i++) {
halfplane L(a.a[nex(i)], a.a[i]);
res = L.cut(res);
}
return res;
}
conv conv_inter(vector<poly> &a) {
conv res;
res.P.push_back(pt(-INF, -INF));
res.P.push_back(pt(INF, -INF));
res.P.push_back(pt(INF, INF));
res.P.push_back(pt(-INF, INF));
for (int cnt = 0; cnt < a.size(); cnt++) {
int n = (int)a[cnt].a.size();
for(int i = 0; i < n; i++) {
halfplane L(a[cnt].a[nex(i)], a[cnt].a[i]);
res = L.cut(res);
}
}
return res;
}
}
int main()
{
int n, m;
scanf("%d", &n);
vector<poly> p(n);
for (int i = 0; i < n; i++) {
scanf("%d", &m);
p[i].a.resize(m);
for(int j = 0; j < m; j++) p[i].a[j].input();
p[i].fix();
}
conv ans = Halfplane::conv_inter(p);
printf("%.3lf\n", ans.area());
return 0;
}