• BZOJ 2618: [Cqoi2006]凸多边形 (半平面交)


    模板

    CODE

    #include <bits/stdc++.h>
    using namespace std;
    const double eps = 1e-10;
    const int MAXN = 505;
    inline double sqr(double x) { return x*x; }
    inline double dcmp(double x) {
        if(x < -eps) return -1;
        if(x > eps) return 1;
        return 0;
    }
    struct Point {
    	double x, y;
    	Point(){}
    	Point(const double &x, const double &y):x(x), y(y){}
    	inline Point operator +(const Point &o)const { return Point(x + o.x, y + o.y); }
    	inline Point operator -(const Point &o)const { return Point(x - o.x, y - o.y); }
    	inline Point operator *(const double &k)const { return Point(x * k, y * k); }
    	inline double operator *(const Point &o)const { return x * o.y - y * o.x; }
    	inline friend double dist(const Point &A, const Point &B) { return sqrt(sqr(A.x-B.x) + sqr(A.y-B.y)); }
    }a[MAXN];
    struct Line {
    	Point p, v; double angle;
    	Line(){}
    	Line(const Point &p, const Point &v):p(p), v(v){
    		angle = atan2(v.y, v.x);
    	}
    	inline friend bool On_Right(const Line &l, const Point &p) {
    		return dcmp(l.v * (p - l.p)) <= 0;
    	}
    	inline bool operator <(const Line &o)const {
    	    if(!dcmp(angle-o.angle))
    	    	return dcmp(v * (o.p - p)) < 0;  //同角度判一下左右关系,必须是严格的判断,否则sort会出错
    	    return angle < o.angle;
        }
    	inline friend Point Get_Intersection(const Line &l1, const Line &l2) {
    		Point u = l1.p - l2.p;
    		double k = (l2.v * u) / (l1.v * l2.v);
    		return l1.p + l1.v * k;
    	}
    }arr[MAXN], q[MAXN]; int head, tail;
    
    inline void Insert(const Line &l) {
    	while(tail-head > 1 && On_Right(l, Get_Intersection(q[tail-2], q[tail-1]))) --tail;
    	while(tail-head > 1 && On_Right(l, Get_Intersection(q[head], q[head+1]))) ++head;
    	q[tail++] = l;
    }
    
    int n, m, tot;
    int main() {
    	scanf("%d", &m);
    	while(m--) {
    		scanf("%d", &n);
    		for(int i = 1; i <= n; ++i) {
    			scanf("%lf%lf", &a[i].x, &a[i].y);
    			if(i > 1) arr[++tot] = Line(a[i-1], a[i]-a[i-1]);
    		}
    		arr[++tot] = Line(a[n], a[1]-a[n]);
    	}
    
    	sort(arr + 1, arr + tot + 1);
    	q[tail++] = arr[1];
    	for(int i = 2; i <= tot; ++i)
            if(dcmp(arr[i].angle-arr[i-1].angle))
                Insert(arr[i]);
    	while(tail-head > 1 && On_Right(q[head], Get_Intersection(q[tail-2], q[tail-1]))) --tail;
    	if(tail - head < 3) return puts("0.000"), 0;
    	
    	Point S = Get_Intersection(q[head], q[tail-1]), last = S;
    	double ans = 0;
    	for(int i = head+1; i < tail; ++i) {
    		Point now = Get_Intersection(q[i], q[i-1]);
    		ans += last * now; last = now;
    	}
    	ans += last * S;
    	printf("%.3f
    ", ans/2);
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Orz-IE/p/12039305.html
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