LA 3263 (平面图的欧拉定理) That Nice Euler Circuit

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.\n", ++kase, e+2-c);

138     }

139 

140     return 0;

141 }

代码君