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  • 分割线算图形面积 圆和多边形面积并

    冬令营听了hwd的课,感觉好像计算几何很简单的样子。

    最近做了一道计算圆和多边形面积并的面积,用了扫面线的方法,感觉不太好(可能我没写习惯),写了(7k)的代码才A。QAQ

    以后记得找月下柠檬树做做。

    代码

    #include <cstdio>
    #include <iostream>
    #include <vector>
    #include <iomanip>
    #include <cmath>
    #include <assert.h>
    using namespace std;
    
    const int MAXN = 103;
    const long double PI = acos(-1);
    const long double EPS = 1e-15;
    
    template <class T>
    T sqr(const T &x) {
    	return x * x;
    }
    
    int n, m;
    
    struct Point {
    	long double x, y;
    
    	Point () {}
    
    	Point (long double a, long double b):x(a), y(b) {}
    
    	Point operator - (const Point &o) const {
    		return Point(x - o.x, y - o.y);
    	}
    
    	Point operator + (const Point &o) const {
    		return Point(x + o.x, y + o.y);
    	}
    	
    	long double operator * (const Point &o) const {
    		return x * o.y - y * o.x;
    	}
    
    	long double operator % (const Point &o) const {
    		return x * o.x + y * o.y;
    	}
    
    	Point operator * (const long double &o) const {
    		return Point(x * o, y * o);
    	}
    
    	Point operator / (const long double &o) const {
    		return Point(x / o, y / o);
    	}
    };
    
    struct Segment {
    	Point first, second;
    
    	Segment() {}
    
    	Segment(Point x, Point y):first(x), second(y) {}
    };
    
    struct Circle {
    	Point center;
    	long double r;
    };
    
    Circle A[MAXN];
    Point B[MAXN];
    
    Point ret0, ret1;
    
    long double dis(const Point &a, const Point &b) {
    	return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
    }
    
    Point rotate(const Point &p, const long double &c, const long double &s) {
    	return Point(p.x * c - p.y * s, p.x * s + p.y * c);
    }
    
    bool cir_cir(const Circle &a, const Circle &b) {
    	long double len = dis(a.center, b.center);
    	if (a.r + b.r < len) return false;
    	if (len < fabs(a.r - b.r)) return false;
    	long double COS = (sqr(a.r) + sqr(len) - sqr(b.r)) / (len * a.r * 2);
    	long double SIN = sqrt(1. - sqr(COS));
    	Point tmp = b.center - a.center;
    	long double k = a.r / len;
    	ret0 = a.center + rotate(tmp, COS, SIN) * k;
    	ret1 = a.center + rotate(tmp, COS, -SIN) * k;
    	return true;
    }
    
    Point line_line(const Segment &a, const Segment &b) {
    	long double k = (a.second - a.first) * (b.first - a.first);
    	k = k / (k - (a.second - a.first) * (b.second - a.first));
    	return b.first + (b.second - b.first) * k;
    }
    
    bool cir_line(const Circle &a, const Segment &b) {
    	Point v = b.second - b.first;
    	swap(v.x, v.y);
    	v.x = -v.x;
    	v = a.center + v;
    	Point A = line_line(Segment(v, a.center), b);
    	long double d = dis(A - a.center, Point(0, 0));
    	if (d > a.r) return false;
    	long double len = sqrt(sqr(a.r) - sqr(d));
    	long double norm = dis(b.second - b.first, Point(0, 0));
    	ret0 = A + (b.second - b.first) * len / norm;
    	ret1 = A - (b.second - b.first) * len / norm;
    	return true;
    }
    
    long double cirSubArea(const Circle &a, const Point &p, const Point &q) {
    	long double X = dis(p, q);
    	long double angle = acos(1. - sqr(X) * .5 / sqr(a.r));
    		//acos((sqr(a.r) * 2 - sqr(X)) * .5 / sqr(a.r));
    	long double ret = angle * sqr(a.r) * .5;
    	ret -= fabs((p - a.center) * (q - a.center) * .5);
    	return ret;
    }
    
    struct Query {
    	int belong;
    	bool enter;
    	Point left, right;
    	long double key;
    
    	bool operator < (const Query &o) const {
    		return key < o.key;
    	}
    };
    
    int main() {
    	freopen("polygon.in", "r", stdin);
    	freopen("polygon.out", "w", stdout);
    
    	cin >> n >> m;
    	vector<long double> key;
    
    	for (int i = 0; i < n; i ++) {
    		cin >> A[i].center.x >> A[i].center.y >> A[i].r;
    		key.push_back(A[i].center.x - A[i].r);
    		key.push_back(A[i].center.x + A[i].r);
    		for (int j = 0; j < i; j ++) {
    			if (cir_cir(A[i], A[j])) {
    				key.push_back(ret0.x);
    				key.push_back(ret1.x);
    			}
    		}
    	}
    	for (int i = 0; i < m; i ++) {
    		cin >> B[i].x >> B[i].y;
    		key.push_back(B[i].x);
    	}
    	B[m] = B[0];
    	for (int i = 0; i < n; i ++) {
    		for (int j = 0; j < m; j ++) {
    			if (cir_line(A[i], Segment(B[j], B[j + 1]))) {
    				if (ret0.x >= min(B[j].x, B[j + 1].x) - EPS &&
    						ret0.x <= max(B[j].x, B[j + 1].x) + EPS)
    					key.push_back(ret0.x);
    				if (ret1.x >= min(B[j].x, B[j + 1].x) - EPS &&
    						ret1.x <= max(B[j].x, B[j + 1].x) + EPS)
    					key.push_back(ret1.x);
    			}
    		}
    	}
    
    	sort(key.begin(), key.end());
    	vector<Query> que;
    
    	long double ans = 0;
    
    	for (int i = 0, _i = key.size(); i + 1 < _i; i ++) {
    		if (i < _i && key[i] == key[i + 1]) continue;
    		long double mid = (key[i] + key[i + 1]) * .5;
    		Segment vecl = Segment(Point(key[i], 0), Point(key[i], 1));
    		Segment vecr = Segment(Point(key[i + 1], 0), Point(key[i + 1], 1));
    		Segment vecm = Segment(Point(mid, 0), Point(mid, 1));
    		que.clear();
    
    		for (int j = 0; j < n; j ++) {
    			if (cir_line(A[j], vecm)) {
    				if (ret0.y > ret1.y) swap(ret0, ret1);
    				long double key0 = ret0.y, key1 = ret1.y;
    
    				cir_line(A[j], vecl);
    				Point a = ret0, b = ret1;
    				if (a.y > b.y) swap(a, b);
    				cir_line(A[j], vecr);
    				if (ret0.y > ret1.y) swap(ret0, ret1);
    				Query tmp;
    				tmp.belong = j;
    
    				tmp.left = a;
    				tmp.right = ret0;
    				tmp.enter = true;
    				tmp.key = key0;
    				que.push_back(tmp);
    				tmp.left = b;
    				tmp.right = ret1;
    				tmp.enter = false;
    				tmp.key = key1;
    				que.push_back(tmp);
    			}
    		}
    		bool in = false;
    		for (int j = 0; j < m; j ++) {
    			Point t = line_line(Segment(B[j], B[j + 1]), vecm);
    			if (((B[j + 1] - B[j]) % (t - B[j]) > 0) == 
    					((B[j + 1] - B[j]) % (t - B[j + 1]) > 0)) continue;
    
    			in = true;
    			Query tmp;
    			tmp.belong = -1;
    			tmp.left = line_line(vecl, Segment(B[j], B[j + 1]));
    			tmp.right = line_line(vecr, Segment(B[j], B[j + 1]));
    			tmp.enter = false;
    			tmp.key = t.y;
    			que.push_back(tmp);
    		}
    		if (in) {
    			int t = que.size();
    			if (que[t - 2].left.y >= que[t - 1].left.y &&
    					que[t - 2].right.y >= que[t - 1].right.y)
    				que[t - 1].enter = true;
    			else
    				que[t - 2].enter = true;
    		}
    
    		sort(que.begin(), que.end());
    
    		int cover = 0;
    		Query lower;
    		for (int j = 0, _j = que.size(); j < _j; j ++) {
    			Query &cur = que[j];
    			if (cur.enter) {
    				if (cover == 0) {
    					lower = cur;
    					if (cur.belong != -1)
    						ans += cirSubArea(A[cur.belong], 
    								cur.left, cur.right);
    				}
    				cover ++;
    			}
    			else {
    				cover --;
    				if (cover == 0) {
    					ans += (cur.left.y + cur.right.y - lower.left.y - lower.right.y) *
    						(key[i + 1] - key[i]) * .5;
    					if (cur.belong != -1) 
    						ans += cirSubArea(A[cur.belong], cur.left, cur.right);
    				}
    			}
    		}
    	}
    
    	//cout << ans << endl;
    	printf("%.15lf
    ", (double) ans);
    
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/wangck/p/4461534.html
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