uva 12296 Pieces and Discs

题意：

有个矩形，左下角(0,0),左上角(L,W）.

思路：

除了圆盘之外，本题的输入也是个PSLG，因此可以按照前面叙述的算法求出各个区域：只需把线段视为直线，用切割凸多边形的方法 ：每次读入线段，切割所有块，最终得到若干凸多边形

如何判断多边形是否与圆盘相交：如果多边形的边和圆周规范相交，圆盘和多边形一定相交，

 1：即使完全没有公共点，也可以相交，互相内含 需要判断多边形是否有顶点在园内，还需要判断圆心是否在多边形内；

2.如果不规范：  a:带判断的线段在园外；

b:带判断的线段在园内即两个端点在圆上，只需要判断中点是否在园内；

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<vector>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long int
#define up(i,x,y) for(i=x;i<=y;i++)
#define w(a) while(a)
const double inf=0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
using namespace std;

double dcmp(double x)
{
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x),y(y) { }
};

typedef Point Vector;

typedef vector<Point> Polygon;

Vector operator + (Vector A, Vector B)
{
    return Vector(A.x+B.x, A.y+B.y);
}

Vector operator - (Point A, Point B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

Vector operator * (Vector A, double p)
{
    return Vector(A.x*p, A.y*p);
}

double Dot(Vector A, Vector B)
{
    return A.x*B.x + A.y*B.y;
}
double Cross(Vector A, Vector B)
{
    return A.x*B.y - A.y*B.x;
}
double Length2(Vector A)
{
    return Dot(A, A);
}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)
{
    Vector u = P-Q;
    double t = Cross(w, u) / Cross(v, w);
    return P+v*t;
}

bool OnSegment(Point p, Point a1, Point a2)
{
    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

// 多边形的有向面积
double PolygonArea(Polygon poly)
{
    double area = 0;
    int n = poly.size();
    for(int i = 1; i < n-1; i++)
        area += Cross(poly[i]-poly[0], poly[(i+1)%n]-poly[0]);
    return area/2;
}

// cut with directed line A->B, return the left part
// may return a single point or a line segment
Polygon CutPolygon(Polygon poly, Point A, Point B)
{
    Polygon newpoly;
    int n = poly.size();
    for(int i = 0; i < n; i++)
    {
        Point C = poly[i];
        Point D = poly[(i+1)%n];
        if(dcmp(Cross(B-A, C-A)) >= 0) newpoly.push_back(C);
        if(dcmp(Cross(B-A, C-D)) != 0)
        {
            Point ip = GetLineIntersection(A, B-A, C, D-C);
            if(OnSegment(ip, C, D)) newpoly.push_back(ip);
        }
    }
    return newpoly;
}

int isPointInPolygon(Point p, Polygon v)
{
    int wn = 0;
    int n = v.size();
    for(int i = 0; i < n; i++)
    {
        if(OnSegment(p, v[i], v[(i+1)%n])) return -1; // 在边界上
        int k = dcmp(Cross(v[(i+1)%n]-v[i], p-v[i]));
        int d1 = dcmp(v[i].y - p.y);
        int d2 = dcmp(v[(i+1)%n].y - p.y);
        if(k > 0 && d1 <= 0 && d2 > 0) wn++;
        if(k < 0 && d2 <= 0 && d1 > 0) wn--;
    }
    if (wn != 0) return 1; // 内部
    return 0; // 外部
}

// 点在圆心内。圆周上不算
bool isInCircle(Point p, Point center, double R)
{
    return dcmp(Length2(p-center) - R*R) < 0;
}

// 直线AB和圆心为C，半径为r的圆的交点
// 返回交点个数，t1, t2分别为两个交点在直线方程中的参数，p1和p2为交点本身
int getLineCircleIntersection(Point A, Point B, Point C, double r, double& t1, double& t2)
{
    // 初始方程：(A.x + t(B.x - A.x) - C.x)^2 + (A.y + t(B.y - A.y) - C.y)^2 = r^2
    // 整理得：(at + b)^2 + (ct + d)^2 = r^2
    double a = B.x - A.x;
    double b = A.x - C.x;
    double c = B.y - A.y;
    double d = A.y - C.y;
    // 展开得：(a^2 + c^2)t^2 + 2(ab + cd)t + b^2 + d^2 - r^2 = 0，即et^2 + ft + g = 0
    double e = a * a + c * c;
    double f = 2 * (a * b + c * d);
    double g = b * b + d * d - r * r;
    double delta = f * f - 4 * e * g; // 判别式
    if(dcmp(delta) < 0) return 0; // 相离
    if(dcmp(delta) == 0)  // 相切
    {
        t1 = t2 = -f / (2 * e);
        return 1;
    }
    t1 = (-f - sqrt(delta)) / (2 * e);
    t2 = (-f + sqrt(delta)) / (2 * e);
    return 2;
}

// 圆和线段是否相交（相切不算）。线段不考虑端点
bool CircleIntersectSegment(Point A, Point B, Point p, double R)
{
    double t1, t2;
    int c = getLineCircleIntersection(A, B, p, R, t1, t2);
    if(c <= 1) return false;
    if(dcmp(t1) > 0 && dcmp(t1-1) < 0) return true; // 端点在圆上
    if(dcmp(t2) > 0 && dcmp(t2-1) < 0) return true;
    return false;
}

/////////// 题目相关
vector<Polygon> pieces, new_pieces;

void cut(int x1, int y1, int x2, int y2)
{
    new_pieces.clear();
    for(int i = 0; i < pieces.size(); i++)
    {
        Polygon left = CutPolygon(pieces[i], Point(x1, y1), Point(x2, y2));
        Polygon right = CutPolygon(pieces[i], Point(x2, y2), Point(x1, y1));
        if(left.size() >= 3) new_pieces.push_back(left);
        if(right.size() >= 3) new_pieces.push_back(right);
    }
    pieces = new_pieces;
}

bool DiscIntersectPolygon(Polygon poly, Point p, double R)
{
    if(isPointInPolygon(p, poly) != 0) return true;
    if(isInCircle(poly[0], p, R)) return true;
    int n = poly.size();
    for(int i = 0; i < n; i++)
    {
        if(CircleIntersectSegment(poly[i], poly[(i+1)%n], p, R))
        {
            return true; // 不考虑线段端点
        }
        if(isInCircle((poly[i]+poly[(i+1)%n])*0.5, p, R))
        {
            return true; // 两个端点都在圆上
        }
    }
    return false;
}

void query(Point p, int R)
{
    vector<double> ans;
    for(int i = 0; i < pieces.size(); i++)
    {
        if(DiscIntersectPolygon(pieces[i], p, R))
        {
            ans.push_back(fabs(PolygonArea(pieces[i])));
        }
    }
    printf("%d", ans.size());
    sort(ans.begin(), ans.end());
    for(int i = 0; i < ans.size(); i++)
        printf(" %.2lf", ans[i]);
    printf("\n");
}

int main()
{
    int n, m, L, W;
    while(scanf("%d%d%d%d", &n, &m, &L, &W) == 4 && n)
    {
        pieces.clear();

        Polygon bbox;
        bbox.push_back(Point(0, 0));
        bbox.push_back(Point(L, 0));
        bbox.push_back(Point(L, W));
        bbox.push_back(Point(0, W));
        pieces.push_back(bbox);

        for(int i = 0; i < n; i++)
        {
            int x1, y1, x2, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            cut(x1, y1, x2, y2);
        }

        for(int i = 0; i < m; i++)
        {
            int x, y, R;
            scanf("%d%d%d", &x, &y, &R);
            query(Point(x, y), R);
        }
        printf("\n");
    }
    return 0;
}