UVALive - 3263
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 Joey‘s Euler machine works exactly like this. The device consists of a pencil touching the paper, and a control center issuing a sequence of instructions. The paper can be viewed as the infinite two-dimensional plane; that means you do not need to worry about 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). 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 N Output For each test case there will be one output line in the format Case x: There are w pieces., where x is the serial number starting from 1. Note: The figures below illustrate the two sample input cases.
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 |
4,
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;
}