HDU 2528 (平面几何 切割多边形 多边形的面积)

题目链接:点击这里

直接求出总的面积和切割完一侧的面积就好了, 检验板子题.

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const double eps = 1e-8;
const double INF = 1e20;
const double pi = acos (-1.0);

int dcmp (double x) {
    if (fabs (x) < eps) return 0;
    return (x < 0 ? -1 : 1);
}
inline double sqr (double x) {return x*x;}

//*************点
struct Point {
    double x, y;
    Point (double _x = 0, double _y = 0):x(_x), y(_y) {}
    void input () {scanf ("%lf%lf", &x, &y);}
    void output () {printf ("%.2f %.2f\n", x, y);}
    bool operator == (const Point &b) const {
        return (dcmp (x-b.x) == 0 && dcmp (y-b.y) == 0);
    }
    bool operator < (const Point &b) const {
        return (dcmp (x-b.x) == 0 ? dcmp (y-b.y) < 0 : x < b.x);
    }
    Point operator + (const Point &b) const {
        return Point (x+b.x, y+b.y);
    }
    Point operator - (const Point &b) const {
        return Point (x-b.x, y-b.y);
    }
    Point operator * (double a) {
        return Point (x*a, y*a);
    }
    Point operator / (double a) {
        return Point (x/a, y/a);
    }
    double len2 () {//返回长度的平方
        return sqr (x) + sqr (y);
    }
    double len () {//返回长度
        return sqrt (len2 ());
    }
    Point change_len (double r) {//转化为长度为r的向量
        double l = len ();
        if (dcmp (l) == 0) return *this;//零向量返回自身
        r /= l;
        return Point (x*r, y*r);
    }
    Point rotate_left () {//顺时针旋转90度
        return Point (-y, x);
    }
    Point rotate_right () {//逆时针旋转90度
        return Point (y, -x);
    }
    Point rotate (Point p, double ang) {//绕点p逆时针旋转ang
        Point v = (*this)-p;
        double c = cos (ang), s = sin (ang);
        return Point (p.x + v.x*c - v.y*s, p.y + v.x*s + v.y*c);
    }
    Point normal () {//单位法向量
        double l = len ();
        return Point (-y/l, x/l);
    }
};

double cross (Point a, Point b) {//叉积

    return a.x*b.y-a.y*b.x;
}
double dot (Point a, Point b) {//点积

    return a.x*b.x + a.y*b.y;
}
double dis (Point a, Point b) {//两个点的距离

    Point p = b-a; return p.len ();
}
double rad_degree (double rad) {//弧度转化为角度

    return rad/pi*180;
}
double rad (Point a, Point b) {//两个向量的夹角

    return fabs (atan2 (fabs (cross (a, b)), dot (a, b)) );
}
bool parallel (Point a, Point b) {//向量平行
    double p = rad (a, b);
    return dcmp (p) == 0 || dcmp (p-pi) == 0;
}

//************直线 线段
struct Line {
    Point s, e;//直线的两个点
    Line () {}
    Line (Point _s, Point _e) : s(_s), e(_e) {}
    //一个点和倾斜角确定直线
    Line (Point p, double ang) {
        s = p;
        if (dcmp (ang-pi/2) == 0) {
            e = s + Point (0, 1);
        }
        else
            e = s + Point (1, tan (ang));
    }
    //ax+by+c=0确定直线
    Line (double a, double b, double c) {
        if (dcmp (a) == 0) {
            s = Point (0, -c/b);
            e = Point (1, -c/b);
        }
        else if (dcmp (b) == 0) {
            s = Point (-c/a, 0);
            e = Point (-c/a, 1);
        }
        else {
            s = Point (0, -c/b);
            e = Point (1, (-c-a)/b);
        }
    }
    void input () {
        s.input ();
        e.input ();
    }
    void adjust () {
        if (e < s) swap (e, s);
    }
    double length () {//求线段长度
        return dis (s, e);
    }
    double angle () {//直线的倾斜角
        double k = atan2 (e.y-s.y, e.x-s.x);
        if (dcmp (k) < 0) k += pi;
        if (dcmp (k-pi) == 0) k -= pi;
        return k;
    }
};
Point line_intersection (Line a, Line v) {//直线交点
    //调用前确保有交点
    double a1 = cross (v.e-v.s, a.s-v.s);
    double a2 = cross (v.e-v.s, a.e-v.s);
    return Point ((a.s.x*a2-a.e.x*a1)/(a2-a1), (a.s.y*a2-a.e.y*a1)/(a2-a1));
}

//*************多边形
double polygon_area (Point *p, int n) {//多边形的有向面积,加上绝对值就是面积
    //n个点
    double area = 0;
    for (int i = 1; i < n-1; i++) {
        area += cross (p[i]-p[0], p[i+1]-p[0]);
    }
    return area/2;
}

int convex_cut (Line u, Point *p, int n, Point *po) {//直线切割多边形左侧
    //返回切割后多边形的数量
    int top = 0;
    for (int i = 0; i < n; i++) {
        int d1 = dcmp (cross (u.e-u.s, p[i]-u.s));
        int d2 = dcmp (cross (u.e-u.s, p[(i+1)%n]-u.s));
        if (d1 >= 0) po[top++] = p[i];
        if (d1*d2 < 0) po[top++] = line_intersection (u, Line (p[i], p[(i+1)%n]));
    }
    return top;
}

#define maxn 22
int n, m;
Point p[maxn], ch[maxn];
Line l;

int main () {
    while (cin >> n && n) {
        for (int i = 0; i < n; i++) p[i].input ();
        l.input ();
        m = convex_cut (l, p, n, ch);
        double ans1 = fabs (polygon_area (ch, m));
        double ans2 = fabs (polygon_area (p, n))-ans1;
        if (ans1 < ans2) swap (ans1, ans2);
        printf ("%.0f %.0f\n", ans1, ans2);
    }
    return 0;
}