题目链接:P4196 [CQOI2006]凸多边形
题意
给定 (n) 个凸多边形,求它们相交的面积。
思路
半平面交
半平面交的模板题。
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
const db eps = 1e-10;
const db pi = acos(-1.0);
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 1e3 + 10;
inline int dcmp(db x) {
if(fabs(x) < eps) return 0;
return x > 0? 1: -1;
}
class Point {
public:
double x, y;
Point(double x = 0, double y = 0) : x(x), y(y) {}
inline void input() {
scanf("%lf%lf", &x, &y);
}
bool operator<(const Point &a) const {
return (!dcmp(x - a.x))? dcmp(y - a.y) < 0: x < a.x;
}
bool operator==(const Point &a) const {
return dcmp(x - a.x) == 0 && dcmp(y - a.y) == 0;
}
db dis2(const Point a) {
return pow(x - a.x, 2) + pow(y - a.y, 2);
}
db dis(const Point a) {
return sqrt(dis2(a));
}
db dis2() {
return x * x + y * y;
}
db dis() {
return sqrt(dis2());
}
Point operator+(const Point a) {
return Point(x + a.x, y + a.y);
}
Point operator-(const Point a) {
return Point(x - a.x, y - a.y);
}
Point operator*(double p) {
return Point(x * p, y * p);
}
Point operator/(double p) {
return Point(x / p, y / p);
}
db dot(const Point a) {
return x * a.x + y * a.y;
}
db cross(const Point a) {
return x * a.y - y * a.x;
}
db ang(Point a) {
return acos((a.dis() * dis()) / dot(a));
}
};
typedef Point Vector;
Point p[maxn], ip[maxn];
class Line {
public:
Point s, e;
db angle;
Line() {}
Line(Point s, Point e) : s(s), e(e) {}
inline void input() {
s.input();e.input();
}
bool operator<(const Line &a) const {
Line l = a;
if(dcmp(angle - l.angle) == 0) {
return l.toLeftTest(s) == 1;
}
return angle < l.angle;
}
void get_angle() {
angle = atan2(e.y - s.y, e.x - s.x);
}
int toLeftTest(Point p) {
if((e - s).cross(p - s) > 0) return 1;
else if((e - s).cross(p - s) < 0) return -1;
return 0;
}
int linecrossline(Line l) {
if(dcmp((e - s).cross(l.e - l.s)) == 0) {
if(dcmp((l.s - e).cross(l.e - s)) == 0) {
return 0;
}
return 1;
}
return 2;
}
Point crosspoint(Line l) {
db a1 = (l.e - l.s).cross(s - l.s);
db a2 = (l.e - l.s).cross(e - l.s);
db x = (s.x * a2 - e.x * a1) / (a2 - a1);
db y = (s.y * a2 - e.y * a1) / (a2 - a1);
if(dcmp(x) == 0) x = 0;
if(dcmp(y) == 0) y = 0;
return Point(x, y);
}
};
Line l[maxn], q[maxn];
db half_plane(int cnt) {
sort(l + 1, l + 1 + cnt);
// for(int i = 1; i <= cnt; ++i) {
// cout << l[i].s.x << " " << l[i].s.y << " ";
// cout << l[i].e.x << " " << l[i].e.y << endl;
// }
int tmp = 1;
for(int i = 2; i <= cnt; ++i) {
if(dcmp(l[i].angle - l[tmp].angle) == 1) l[++tmp] = l[i];
}
cnt = tmp;
int head = 1, tail = 2;
q[1] = l[1], q[2] = l[2];
for(int i = 3; i <= cnt; ++i) {
while(head < tail && l[i].toLeftTest(q[tail].crosspoint(q[tail - 1])) == -1) {
--tail;
}
while(head < tail && l[i].toLeftTest(q[head].crosspoint(q[head + 1])) == -1) {
++head;
}
q[++tail] = l[i];
}
while(head < tail && q[head].toLeftTest(q[tail].crosspoint(q[tail - 1])) == -1) {
--tail;
}
while(head < tail && q[tail].toLeftTest(q[head].crosspoint(q[head + 1])) == -1) {
++head;
}
if(tail - head + 1 <= 2) {
return 0.0;
}
tmp = 0;
for(int i = head; i < tail; ++i) {
ip[++tmp] = q[i].crosspoint(q[i + 1]);
}
ip[++tmp] = q[head].crosspoint(q[tail]);
db ans = 0;
for(int i = 3; i <= tmp; ++i) {
ans += (ip[i - 1] - ip[1]).cross(ip[i] - ip[1]);
}
return ans * 0.5;
}
int main() {
int n;
scanf("%d", &n);
int cnt = 0;
for(int i = 1; i <= n; ++i) {
int m;
scanf("%d", &m);
for(int i = 0; i < m; ++i) {
p[i].input();
if(i) {
l[++cnt].e = p[i];
l[cnt].s = p[i - 1];
l[cnt].get_angle();
}
}
l[++cnt].e = p[0];
l[cnt].s = p[m - 1];
l[cnt].get_angle();
}
printf("%.3lf
", half_plane(cnt));
return 0;
}