3263 - That Nice Euler CircuitTime limit: 3.000 seconds |
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.
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
if the pencil will ever go off the boundary.
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 N >= 4, 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
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
题目大意
平面上给出n个点，求划分出的平面区域个数
解题思路
欧拉定理：设平面图的顶点数、边数和面数分别是V,E,F，则有V+F-E=2
●顶点个数v:n个顶点+交点，注意存在三线共点的情况，找到所有交点后要去重，利用去重函数unique去重函数unique
●棱数e：n条棱，若有交点棱数要多，注意存在三线共点的情况，并不是简单地一个交点就多以条棱，而是凡是穿过交点的边棱数都加+，所有要遍历边判断交点是否在边上，满足就e++
参考《算法竞赛入门经典——训练指南》第四章 计算几何
参考代码+部分注释
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <climits>
#define eps 1e-10
using namespace std;
typedef long long ll;
const int INF=INT_MAX;
const int maxn = 300+10;
int dcmp(double x){//三态函数，克服浮点数精度陷阱，判断x==0?x<0?x>0?
if(fabs(x)<eps) return 0;else return x<0?-1:1;
}
struct Point{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}//构造函数，方便代码编写
};
typedef Point Vector;//Vector是 Point的别名
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);}
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;}
//向量旋转，rad是弧度不是角度
Vector Rotate(Vector A,double rad){return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}
//直线交点，调用前请确保两条直线P+tv,Q+tw有唯一交点。当且仅当Cross（v,w）非0
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;
}
//线段规范相交判定
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;
}
//点在线上（三点共线）判定
bool OnSegment(Point p,Point a1,Point a2){
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}
Point P[maxn],V[maxn*maxn];//P[]存储n个点,v[]存储可能的顶点
int main()
{
// freopen("input.txt","r",stdin);
int n,kase=0;
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--; //n个顶点，其中有一个重复，要减掉
//求顶点个数v,初始为n,考虑到可能有三线共点的情况，求出所有交点再去重
int v=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++]=GetLineIntersection(P[i],P[i+1]-P[i],P[j],P[j+1]-P[j]);
sort(V,V+v);
v=unique(V,V+v)-V;//去重过程，先排序；最终得到顶点数v
//再求棱e,初始化e=n，注意三线共点的情况，并不是有几个交点棱就+几
int e=n;
for(int i=0;i<v;i++)
for(int j=0;j<n;j++)
if(OnSegment(V[i],P[j],P[j+1])) e++;//遍历边，点再线上说明出现交点，e++
printf("Case %d: There are %d pieces.\n",++kase,e+2-v);
}
return 0;
}