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;}
就微笑着同清风合力染绿大地，这样才算善待生命，不负年华。
