UVALive - 3263 - That Nice Euler Circuit （计算几何~~）
| UVALive - 3263
That Nice Euler Circuit
Description Little Joey invented a scrabble machine that he called Euler, after the great mathematician. In his primary school Joey heard about the nice story of how Euler started the study about graphs. The problem in that story was - let me remind you - to draw a graph on a paper without lifting your pen, and finally return to the original position. Euler proved that you could do this if and only if the (planar) graph you created has the following two properties: (1) The graph is connected; and (2) Every vertex in the graph has even degree.
In the beginning, the Euler machine will issue an instruction of the form (X0, Y0) which moves the pencil to some starting position (X0, Y0). Each subsequent instruction is also of the form (X', Y'), which means to move the pencil from the previous position to the new position (X', Y'), thus draw a line segment on the paper. You can be sure that the new position is different from the previous position for each instruction. At last, the Euler machine will always issue an instruction that move the pencil back to the starting position (X0, Y0). In addition, the Euler machine will definitely not draw any lines that overlay other lines already drawn. However, the lines may intersect. After all the instructions are issued, there will be a nice picture on Joey's paper. You see, since the pencil is never lifted from the paper, the picture can be viewed as an Euler circuit. Your job is to count how many pieces (connected areas) are created on the paper by those lines drawn by Euler. Input There are no more than 25 test cases. Ease case starts with a line containing an integer N4, which is the number of instructions in the test case. The following N pairs of integers give the instructions and appear on a single line separated by single spaces. The first pair is the first instruction that gives the coordinates of the starting position. You may assume there are no more than 300 instructions in each test case, and all the integer coordinates are in the range (-300, 300). The input is terminated when N is 0. Output For each test case there will be one output line in the format
Sample Input 5 0 0 0 1 1 1 1 0 0 0 7 1 1 1 5 2 1 2 5 5 1 3 5 1 1 0 Sample Output Case 1: There are 2 pieces. Case 2: There are 5 pieces. Source
Regionals 2004 >> Asia - Shanghai
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首先，这里要运用到离散数学里的定理——欧拉定理：在平面图中，其顶点，边，面的关系为 v + r - e = 2 （v为顶点数，r为面数，e为边数）
则只需求出顶点数以及边数就可以求出面数了
这里平面图的结点由原来的结点和新增的结点组成，由于可能出现三线共点，需要删除重复的点（这里用unique）
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
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); }
bool operator < (const Point& a, const Point& b) {
return a.x < b.x || (a.x == b.x && a.y < b.y);
}
const double eps = 1e-10;
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) );
}
Vector Normal(Vector A) {
double L = Length(A);
return Vector(-A.y/L, A.x/L);
}
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;
}
double DistanceToLine(Point P, Point A, Point B) {
Vector v1 = B-A, v2 = P - A;
return fabs(Cross(v1,v2) / Length(v1));
}
double DistanceToSegment(Point P, Point A, Point B) {
if(A==B) return Length(P-A);
Vector v1 = B - A, v2 = P - A, v3 = P - B;
if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
else return fabs(Cross(v1, v2)) / Length(v1);
}
Point GetLineProjection(Point P, Point A, Point B) {
Vector v = B - A;
return A + v * ( Dot(v, P-A) / Dot(v, v) );
}
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
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;
}
double ConvexPolygonArea(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 / 2;
}
const int maxn = 300 + 10;
Point P[maxn], V[maxn*maxn];
int main() {
int n, cas = 1;
while(scanf("%d", &n) == 1 && n) {
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(SegmentProperIntersection(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;//unique为去重函数，即“去除”相邻的重复元素，返回值为最后一个顶点地址
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", cas++, e + 2 - c);
}
return 0;
}