题解
- 欧拉定理:顶点数V,边数E,面数F满足V + F - E = 2,所以F = E + 2 - V
- 点数包括原有的点和新的相交的点。所以每两条线段求交点。但是去掉交点重复的点。求得V
- 在原先边的基础上,如果某个点在某条表上,则边数加1。求得E。
- 注意要用unique,结构体要重载 == 运算符。
代码
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
int const N = 300 + 10;
double const eps = 1e-8;
int n,e,c;
int dcmp(double x){ //判断符号
if(fabs(x) < eps) return 0;
else return x < 0 ? -1 : 1;
}
typedef struct Point{ //点和向量
double x,y;
Point(){};
Point(double x,double y):x(x),y(y){};
Point operator - (const Point& e)const{ //减
return Point(x - e.x,y - e.y);
}
Point operator + (const Point& e)const{ //减
return Point(x + e.x,y + e.y);
}
double operator ^ (const Point& e)const{ //叉乘
return x * e.y - y * e.x;
}
double operator * (const Point& e)const{ //点积
return x * e.x + y * e.y;
}
bool operator < (const Point& e)const{ //sort操作
return x < e.x || (x == e.x && y < e.y);
}
bool operator == (const Point& e)const{ //unique操作
return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0;
}
}Vector;
struct Line{ //直线的定义
Point a,b;
Line(){};
Line(Point a,Point b):a(a),b(b){}
};
vector<Point>v;
Point p[N];
bool onsegment(Point p,Point a1,Point a2){ //点p是否在线段a1a2上,端点不重合。如果否则取等号,则端点可以重合。
return dcmp((a1 - p) ^ (a2 - p)) == 0 && dcmp((a1 - p) * (a2 - p)) < 0;
}
bool segment_intersection(Line line1,Line line2){ //判断线段是否相交,相交返回true
double c1 = (line1.b - line1.a) ^ (line2.a - line1.a);
double c2 = (line1.b - line1.a) ^ (line2.b - line1.a);
double c3 = (line2.b - line2.a) ^ (line1.a - line2.a);
double c4 = (line2.b - line2.a) ^ (line1.b - line2.a);
if(dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0) return true;
return false;
}
Point intersection(Point p,Vector v,Point q,Vector w){ //两直线(p+tv)和(q+tw)求交点
Vector u = p - q;
double t = (w ^ u) / (v ^ w);
return p + Point(v.x * t,v.y * t);
}
int main(){
int caser = 0;
while(~scanf("%d",&n) && n){
v.clear();
for(int i=0;i<n;i++){
scanf("%lf%lf",&p[i].x,&p[i].y);
v.push_back(p[i]);
}
e = --n;
for(int i=0;i<n;i++){ //(i,i+1)与(j,j+1)的交点
for(int j=i+1;j<n;j++)
if(segment_intersection(Line(p[i],p[i+1]),Line(p[j],p[j+1])))
v.push_back(intersection(p[i],p[i+1]-p[i],p[j],p[j+1]-p[j]));
}
sort(v.begin(),v.end());
v.erase(unique(v.begin(),v.end()),v.end());
for(int i=0;i<v.size();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",++caser,e + 2 - v.size());
}
return 0;
}