【GDCPC H】【HDU5243】Homework

给出平面上的N个点
对于平面上的一点p，要求任何一条通过p且不经过上面N个点的直线，其两侧的点数都不少于⌊N/3⌋
问符合上面要求的点形成的区域的面积是多少

假如一个点一侧有少于n/3个点，这个点这一侧所有点都不可能

因此对于所有右侧恰好有n/3 - 1个点的直线做半平面交就是答案

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define V P
const double eps = 1e-6;
inline int dcmp (double x) {
return x < -eps ? -1 : x > eps;
}
struct P {
double x, y;
void scan() {
scanf("%lf%lf", &x, &y);
}
P(double _x = 0, double _y = 0) : x(_x), y(_y) { }
V operator + (V a) const {
return V(x + a.x, y + a.y);
}
V operator - (V a) const {
return V(x - a.x, y - a.y);
}
V operator * (double p) const {
return V(p * x, p * y);
}
V operator / (double p) const {
return V(x / p, y / p);
}
bool operator < (P a) const {
return x < a.x || (dcmp(x - a.x) == 0 && y < a.y);
}
bool operator == (P a) const {
return dcmp(x - a.x) == 0 && dcmp(y - a.y) == 0;
}
};

inline double dot(V a, V b) {
return a.x * b.x + a.y * b.y;
}
inline double len(V a) {
return sqrt(dot(a, a));
}
inline double dis(P a, P b) {
return len(b - a);
}
inline double ang(V a, V b) {
return acos(dot(a, b) / len(a) / len(b));
}
inline double cross(V a, V b) {
return a.x * b.y - a.y * b.x;
}
inline double area(P a, P b, P c) {
return cross(b - a, c - a);
}
V rot(V a, double p) {
return V(a.x * cos(p) - a.y * sin(p), a.x * sin(p) + a.y * cos(p));
}
V normal(V a) {
double L = len(a);
return V(-a.y / L, a.x / L);
}
P inter(P p, V v, P q, V w) {
V u = p - q;
double t = cross(w, u) / cross(v, w);
return p + v * t;
}
double dis(P p, P a, P b) {
V v1 = b - a, v2 = p - a;
return fabs(cross(v1, v2)) / len(v1);
}
double dis2(P p, P a, P b) {
if (a == b) return len(p - a);
V v1 = b - a, v2 = p - a, v3 = p - b;
if (dcmp(dot(v1, v2)) < 0) return len(v2);
else if (dcmp(dot(v1, v3)) > 0) return len(v3);
else return fabs(cross(v1, v2)) / len(v1);
}
P proj(P p, P a, P b) {
V v = b - a;
return a + v * (dot(v, p - a) / dot(v, v));
}
bool isInter(P a1, P a2, P b1, P b2) {
double c1 = cross(a2 - a1, b1 - a1), c2 = cross(a2 - a1, b2 - a1),
c3 = cross(b2 - b1, a1 - b1), c4 = cross(b2 - b1, a2 - b1);
return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
bool onSeg(P p, P a1, P a2) {
return dcmp(cross(a1 - p, a2 - p)) == 0 && dcmp(dot(a1 - p, a2 - p)) < 0;
}

double area(P* p, int n) {
double s = 0;
p
= p[0];
for (int i = 1; i < n; i ++)
s += cross(p[i] - p[0], p[i + 1] - p[0]);
return s / 2;
}
int graham(P* p, int n, P* ch) {
sort(p, p + n);
int m = 0;
for (int i = 0; i < n; i ++) {
while (m > 1 && cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m --;
ch[m ++] = p[i];
}
int k = m;
for (int i = n - 2; i >= 0; i --) {
while (m > k && cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m --;
ch[m ++] = p[i];
}
if (n > 1) m --;
return m;
}
struct L {
P p;
V v;
double ang;
L() {}
L(P _p, V _v) : p(_p), v(_v) { ang = atan2(v.y, v.y); }
bool operator < (const L& L) const {
return ang < L.ang;
}
};
inline int get(P a) {
if( a.x > 0 && a.y >= 0) return 1;
if( a.x <= 0 && a.y > 0) return 2;
if( a.x < 0 && a.y <= 0) return 3;
if( a.x >= 0 && a.y < 0) return 4;
return 0;
}
inline bool cmp2 (L a, L b) {
return get(a.v) < get(b.v) || (get(a.v) == get(b.v) && dcmp( cross(a.v, b.v) ) >0);
}
bool onLeft(L l, P p) {
return cross(l.v, p - l.p) > 0;
}
P inter(L a, L b) {
return inter(a.p, a.v, b.p, b.v);
}
int half(L* l, int n, P* po) {
sort(l, l + n, cmp2);
int h, t;
P *p = new P
;
L *q = new L
;
q[h = t = 0] = l[0];
for (int i = 1; i < n; i ++) {
while (h < t && !onLeft(l[i], p[t - 1])) t --;
while (h < t && !onLeft(l[i], p[h])) h ++;
q[++ t] = l[i];
if (dcmp(cross(q[t].v, q[t - 1].v)) == 0) {
t --;
if (onLeft(q[t], l[i].p)) q[t] = l[i];
}
if (h < t) p[t - 1] = inter(q[t - 1], q[t]);
}
while (h < t && !onLeft(q[h], p[t - 1])) t --;
if (t - h <= 1) return 0;
p[t] = inter(q[t], q[h]);
int m = 0;
for (int i = h; i <= t; i ++) po[m ++] = p[i];
return m;
}
inline bool cmp (V a, V b) {
return get(a) < get(b) || (get(a) == get(b) && dcmp( cross(a, b) ) >0);
}
const int N = 101000;
int n;
P a
, b
, res
;
L l
;
int main() {
freopen("a.in", "r", stdin);
int T;
scanf("%d", &T);
for (int cas = 1; cas <= T; cas ++) {
scanf("%d", &n);
for (int i = 0; i < n; i ++)
a[i].scan();
int bound = n / 3 - 1, l_c = 0;
for (int i = 0; i < n; i ++) {
int cnt = 0;
for (int j = 0; j < n; j ++)
if (j != i)
b[cnt ++] = a[j] - a[i];
sort(b, b + cnt, cmp);
int t = 0, sum = 0;
for (int j = 0; j < cnt; j ++) {
//if (j == 1) printf("fuck %d %d\n", dcmp(cross(b[j], b[(t + 1) % cnt])), dcmp(dot(b[j], b[(t + 1) % cnt])));
while ((dcmp(cross(b[j], b[(t + 1) % cnt])) == 1) ||
(dcmp(cross(b[j], b[(t + 1) % cnt])) == 0 &&
dcmp(dot(b[j], b[(t + 1) % cnt])) == -1)) t = (t + 1) % cnt, sum ++;
if (cnt - (sum + 1) == bound) l[l_c ++] = L(a[i], b[j]);
//if (i == 0) printf("%d\n", sum);
if (t == j) t ++;
else {
while (dcmp(cross(b[j], b[(j + 1) % cnt])) == 0) j ++, sum --;
sum --;
}
}
}
//printf("%d\n", l_c);
//for (int i = 0; i < l_c; i ++)
//    printf("%lf %lf %lf %lf\n", l[i].p.x, l[i].p.y, l[i].v.x, l[i].v.y);
int ans = half(l, l_c, res);
//printf("%d\n", ans);
//for (int i = 0; i < ans; i ++)
//    printf("%lf %lf\n", res[i].x, res[i].y);
printf("Case #%d: %.6lf\n", cas, area(res, ans));
}
return 0;
}