UVALA 3263 Audiophobia

点击打开题目链接

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

const double eps = 1e-10;
struct Point        //定义点
{
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}    //构造函数，方便代码编写
};


const int MAXN = 300 + 10;
Point P[MAXN], V[MAXN * MAXN];

typedef Point Vector;       //向量是点的一个别名
//重载 +, -, *, <, == 运算符
Vector operator + (Vector A, Vector B)    //“+”运算符，向量+向量=向量
{
    return Vector(A.x + B.x, A.y + B.y);
}

Vector operator - (Point A, Point B)        //“-”运算符，点-点=向量
{
    return Vector(A.x - B.x, A.y - B.y);
}

Vector operator * (Vector A, double p)      //“*”运算符，向量*数=向量
{
    return Vector(A.x * p, A.y * p);
}

bool operator < (const Point& a, const Point& b)    //“<”运算符
{
    return a.x < b.x || (a.x == b.x && a.y > b.y);
}
//“三态函数”，减少精度问题
int dcmp(double x)
{
    if (fabs(x) < eps) return 0;
    return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point& b)   //“==”运算符
{
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
//点积 等于两向量长度的乘积再乘上它们夹角的余弦
double Dot(Vector A, Vector B)
{
    return A.x * B.x + A.y * B.y;
}
//叉积，两向量组成的三角形的有向面积的两倍
double Cross(Vector A, Vector B)
{
    return A.x * B.y - A.y * B.x;
}
//直线交点，调用前确保两条直线P+tv和Q+tw有唯一交点。当且仅当Cross(v,w)非0
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;
}
//判断线段是否相交，两线段恰好有一个公共点，且不在任意一条线段的端点
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)
{
    double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
           c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
//判断一个点是否在一条线段上（不包含端点）
bool OnSegment(Point p, Point a1, Point a2)
{
    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}

int main()
{
    int n, kcas = 0;
    while (~scanf("%d", &n), n)
    {
        for (int i = 0 ; i < n; i++)
        {
            scanf("%lf%lf", &P[i].x, &P[i].y);
            V[i] = P[i];
        }
        n--;
        int numOfPoint = n, numOfEdge = n;
        for (int i = 0; i < n; i++)         //找出所有线段相交形成的新增点
            for (int j = i + 1; j < n; j++)
                if (SegmentProperIntersection(P[i], P[i + 1], P[j], P[j + 1]))
                    V[numOfPoint++] = GetLineIntersection(P[i], P[i + 1] - P[i], P[j], P[j + 1] - P[j]);

        sort(V, V + numOfPoint);            //排序
        numOfPoint = unique(V, V + numOfPoint) - V; //删除重复的点
        for (int i = 0; i < numOfPoint; i++)
            for (int j = 0; j < n; j++)
                if (OnSegment(V[i], P[j], P[j + 1])) numOfEdge++;
        printf("Case %d: There are %d pieces.\n", ++kcas, numOfEdge + 2 - numOfPoint);
    }
    return 0;
}