【poj 2451】半平面交_枚举半平面交c语言-优快云博客

本文介绍了一种基于几何图形的面积计算算法实现过程，包括数据结构定义、算法设计及优化，通过实例演示了如何利用特定算法求解不同几何图形的面积。

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注意先弹出队顶，dz < 3 的特判

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for(int i = pos[x]; i; i = g[i].nex)
#define eps 0.00000001
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 20005;
struct node { DB x, y; } d1, d2, d4, d[N], O;
struct line {
	node u, v; DB rad;
	void Set() { rad = atan2(v.y - u.y, v.x - u.x); }
} a[N], q[N];
int n, hd, tl, dz; DB ans;
int D(DB x) {
	if (x > eps) return 1;
	return (x > -eps) ? 0 : -1;
}
DB Cross(node o, node a, node b) { return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x); }
bool cmp(line a, line b) {
	return (D(a.rad - b.rad) == 0) ? Cross(a.u, a.v, b.u) < 0 : a.rad < b.rad;
}
node P(line a, line b) {
	DB k1 = Cross(a.u, a.v, b.u), k2 = Cross(a.u, a.v, b.v), t;
	t = k1 / (k1 - k2);
	node c;
	c.x = b.u.x + t * (b.v.x - b.u.x);
	c.y = b.u.y + t * (b.v.y - b.u.y);
	return c;
}
void Build() {
	q[hd = tl = 1] = a[1];
	Rep(i, 2, n) {
		if (D(a[i - 1].rad - a[i].rad) == 0) continue ;
		while (hd < tl && Cross(a[i].u, a[i].v, P(q[tl], q[tl - 1])) <= 0) tl --;
		while (hd < tl && Cross(a[i].u, a[i].v, P(q[hd], q[hd + 1])) <= 0) hd ++;
		q[++ tl] = a[i];
	}
	while (hd < tl && Cross(q[hd].u, q[hd].v, P(q[tl], q[tl - 1])) <= 0) tl --;
	while (hd < tl && Cross(q[tl].u, q[tl].v, P(q[hd], q[hd + 1])) <= 0) hd ++;
	Rep(i, hd + 1, tl) d[++ dz] = P(q[i], q[i - 1]);
	if (hd < tl - 1) d[++ dz] = P(q[hd], q[tl]);
}
void Area() {
	if (dz < 3) return ;
	Rep(i, 1, dz) ans += Cross(O, d[i], d[i + 1]);
	ans += Cross(O, d[dz], d[1]);
}
int main()
{
	scanf ("%d", &n);
	Rep(i, 1, n) {
		scanf ("%lf%lf%lf%lf", &a[i].u.x, &a[i].u.y, &a[i].v.x, &a[i].v.y);
	}
	d1.x = d4.x = d1.y = d2.y = 10000, d2.x = d4.y = 0;
	a[n + 1].u = d1, a[n + 1].v = d2;
	a[n + 2].u = d2, a[n + 2].v = O;
	a[n + 3].u = O, a[n + 3].v = d4;
	a[n + 4].u = d4, a[n + 4].v = d1;
	n += 4;
	Rep(i, 1, n) a[i].Set();
	sort(a + 1, a + n + 1, cmp);
	Build(); Area();
	printf("%.1f\n", fabs(ans) / 2);

	return 0;
}