题意：
平面上有n个端点的一笔画，最后一个端点与第一个端点重合，即所给图案是闭合曲线。求这些线段将平面分成多少部分。
分析：
平面图中欧拉定理：设平面的顶点数、边数和面数分别为V、E和F。则 V+F-E=2
所求结果不容易直接求出，因此我们可以转换成 F=E-V+2
枚举两条边，如果有交点则顶点数+1，并将交点记录下来
所有交点去重(去重前记得排序)，如果某个交点在线段上，则边数+1
1 //#define LOCAL 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 using namespace std; 7 8 const int maxn = 300 + 10; 9 10 struct Point 11 { 12 double x, y; 13 Point(double x=0, double y=0) :x(x),y(y) {} 14 }; 15 typedef Point Vector; 16 const double EPS = 1e-10; 17 18 Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } 19 20 Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); } 21 22 Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } 23 24 Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } 25 26 bool operator < (const Point& a, const Point& b) 27 { return a.x < b.x || (a.x == b.x && a.y < b.y); } 28 29 int dcmp(double x) 30 { if(fabs(x) < EPS) return 0; 31 else return x < 0 ? -1 : 1; } 32 33 bool operator == (const Point& a, const Point& b) 34 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; } 35 36 double Dot(Vector A, Vector B) 37 { return A.x*B.x + A.y*B.y; } 38 39 double Length(Vector A) { return sqrt(Dot(A, A)); } 40 41 double Angle(Vector A, Vector B) 42 { return acos(Dot(A, B) / Length(A) / Length(B)); } 43 44 double Cross(Vector A, Vector B) 45 { return A.x*B.y - A.y*B.x; } 46 47 double Area2(Point A, Point B, Point C) 48 { return Cross(B-A, C-A); } 49 50 Vector VRotate(Vector A, double rad) 51 { 52 return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); 53 } 54 55 Point PRotate(Point A, Point B, double rad) 56 { 57 return A + VRotate(B-A, rad); 58 } 59 60 Vector Normal(Vector A) 61 { 62 double l = Length(A); 63 return Vector(-A.y/l, A.x/l); 64 } 65 66 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) 67 { 68 Vector u = P - Q; 69 double t = Cross(w, u) / Cross(v, w); 70 return P + v*t; 71 } 72 double DistanceToLine(Point P, Point A, Point B) 73 { 74 Vector v1 = B - A, v2 = P - A; 75 return fabs(Cross(v1, v2)) / Length(v1); 76 } 77 78 double DistanceToSegment(Point P, Point A, Point B) 79 { 80 if(A == B) return Length(P - A); 81 Vector v1 = B - A, v2 = P - A, v3 = P - B; 82 if(dcmp(Dot(v1, v2)) < 0) return Length(v2); 83 else if(dcmp(Dot(v1, v3)) > 0) return Length(v3); 84 else return fabs(Cross(v1, v2)) / Length(v1); 85 } 86 87 Point GetLineProjection(Point P, Point A, Point B) 88 { 89 Vector v = B - A; 90 return A + v * (Dot(v, P - A) / Dot(v, v)); 91 } 92 93 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) 94 { 95 double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1); 96 double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1); 97 return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0; 98 } 99 100 bool OnSegment(Point P, Point a1, Point a2) 101 { 102 Vector v1 = a1 - P, v2 = a2 - P; 103 return dcmp(Cross(v1, v2)) == 0 && dcmp(Dot(v1, v2)) < 0; 104 } 105 106 Point P[maxn], V[maxn*maxn]; 107 108 int main(void) 109 { 110 #ifdef LOCAL 111 freopen("3263in.txt", "r", stdin); 112 #endif 113 114 int n, kase = 0; 115 while(scanf("%d", &n) == 1 && n) 116 { 117 for(int i = 0; i < n; ++i) 118 { 119 scanf("%lf%lf", &P[i].x, &P[i].y); 120 V[i] = P[i]; 121 } 122 n--; 123 int c = n, e = n; 124 125 for(int i = 0; i < n; ++i) 126 for(int j = i+1; j < n; ++j) 127 if(SegmentProperIntersection(P[i], P[i+1], P[j], P[j+1])) 128 V[c++] = GetLineIntersection(P[i], P[i+1]-P[i], P[j], P[j+1]-P[j]); 129 130 sort(V, V+c); 131 c = unique(V, V+c) - V; 132 133 for(int i = 0; i < c; ++i) 134 for(int j = 0; j < n; ++j) 135 if(OnSegment(V[i], P[j], P[j+1])) e++; 136 137 printf("Case %d: There are %d pieces. ", ++kase, e+2-c); 138 } 139 140 return 0; 141 }