UVA 1342 That Nice Euler Circuit

大意：平面上有n个端点的一笔画，第n个端点总是和第一个端点重合。按照顺序给你一些点，然后通过铅笔的移动跟随点的移动从而画出移动的痕迹，保证后一条线段不覆盖前一条线段，问这些线段将平面分成多少部分？

思路：

欧拉定理：平面顶点数V、边数E和面数F，则V+F-E = 2;

这样只需求出顶点数V和边数E，这样就可以求出F = E+2-V。

设平面的结点由两部分组成，即原来的结点和新增的结点，由于可能出现三点共线，需要删除重复的节点。

由题可知，先前的结点数为n，边数也为n，这样，通过线段判交找出新增的结点（不算两端），然后算出新增的节点，然后判断新增的结点有多少在线段上（不算两端），则新增加了多少新边。

#include <iostream>#include <cstdlib>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <stack>#include <algorithm>using namespace std;const double eps = 1e-10;struct Point{double x, y;Point(double x = 0, double y = 0) : x(x), y(y) { }bool operator < (const Point& a) const{if(a.x != x) return x < a.x;return y < a.y;}};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }Vector operator - (Point A, Point 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); }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 Length(Vector A) { return sqrt(Dot(A, A)); }double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }Vector Rotate(Vector A, double rad){return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));}Point GetIntersection(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 SegmentProperIntersection(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;}double PolygonArea(Point* p, int n){double area = 0;for(int i = 1; i < n-1; i++)area += Cross(p[i]-p[0], p[i+1]-p[0]);return area;}bool OnSegment(Point p, Point a1, Point a2){return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;}Point read_point(){Point A;scanf("%lf%lf", &A.x, &A.y);return A;}int n;const int maxn = 310;Point P[maxn], V[maxn*maxn];int read_case(){scanf("%d", &n);if(!n) return 0; for(int i = 0; i < n; i++) P[i] = read_point(), V[i] = P[i];n--;return 1;}void solve(){int v = n, e = n;for(int i = 0; i < n; i++){for(int j = i+1; j < n; j++){if(SegmentProperIntersection(P[i], P[i+1], P[j], P[j+1])){V[v++] = GetIntersection(P[i], P[i+1]-P[i], P[j], P[j+1]-P[j]); //新增加的点数 }}}sort(V, V+v); //排序 v = unique(V, V+v) - V; //去重 for(int i = 0; i < v; i++){for(int j = 0; j < n; j++) //判断新增加的边数 {if(OnSegment(V[i], P[j], P[j+1])) e++;}}int ans = e+2-v;printf("There are %d pieces.\n", ans);}int main(){int times = 0;while(read_case()){printf("Case %d: ", ++times);solve();}return 0;}