UVALA 3263 That Nice Euler Circuits(欧拉定理，判断线段相交）

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解题思路：

欧拉定理：设平面图的顶点数，边数和面数分别为V，E， F则 V + F - E = 2；

本题要求平面数，即 F = E + 2 - V；

因此只需要求出顶点数和边数。顶点数除了输入的顶点还包括两条线段相交的交点，同样如果三点共线，则原来的一条边变成了两条边。

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <map>#define LL long longusing namespace std;const int MAXN = 300 + 10;struct Point{    double x, y;    Point (double x = 0, double y = 0) : x(x), y(y) { }};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }Vector operator - (Vector A, Vector 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); }Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }const double eps = 1e-10;bool operator <(const Point& a, const Point& b){    return a.x < b.x || (a.x == b.x && a.y < b.y);}int dcmp(double x){    if(fabs(x) < eps) return 0;    else return x < 0 ? -1 : 1;}bool operator == (const Point& a, const Point& b){    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;}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;}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 SegmentIntersection(Point a1, Point a2, Point b1, Point b2){    double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1);    double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;}bool OnSegment(Point p, Point a1, Point a2){    return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;}Point P[MAXN], V[MAXN*MAXN];int main(){    int n, kcase = 1;    while(scanf("%d", &n)!=EOF)    {        if(n == 0) break;        for(int i=0;i<n;i++)        {            scanf("%lf%lf", &P[i].x, &P[i].y);            V[i] = P[i];        }        n--;        int c = n, e = n;        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                if(SegmentIntersection(P[i], P[i+1], P[j], P[j+1]))                   V[c++] = GetLineIntersection(P[i], P[i+1]-P[i], P[j], P[j+1]-P[j]);            }        }        sort(V, V+c);        c = unique(V, V+c) - V;        for(int i=0;i<c;i++)        {            for(int j=0;j<n;j++)            {                if(OnSegment(V[i], P[j], P[j+1])) e++;            }        }        printf("Case %d: There are %d pieces.\n", kcase++, e + 2 - c);    }    return 0;}

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