LA3263 一笔画

题目大意：
依次给定多个点（要求第一个点和最后一个点重叠），把前后两个点相连求最后得到的图形的面的个数

根据欧拉定理：

设平面图的顶点数为V，边数为E，面数为F，则V+F-E = 2

这里的E是指如果一条直线上被多个点分割，那么就算多条边

所以我们要求出V和E的值

先求点，已给定的点数，还要包括相连过程中相交得到的点，经过去重得到最后的点数

for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                //if(i==j) continue;
                if(onIntersection(po[i],po[i+1],po[j],po[j+1])){
                    vp.push_back(getLineIntersection(po[i],po[i+1]-po[i],po[j],po[j+1]-po[j]));
                }
            }
        }
        sort(vp.begin(),vp.end());
        c=unique(vp.begin(),vp.end()) - vp.begin(); //去重前要先进行排序，unique是对地址进行操作，所以这里使用数组也可以

然后找边，根据是否有新得到的点出现在边上，若有，每次边数++；

for(int i=0;i<c;i++){
            for(int j=0;j<n;j++){
                if(onSegment(vp[i],po[j],po[j+1]))
                    e++;
            }
        }

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define eps 1e-10
 struct Point {
     double x,y;
     Point(double x=,double y=):x(x),y(y){}
 }po[];
 typedef Point Vector;

 vector<Point> vp;

 int dcmp(double x)
 {
     if(abs(x)<eps) return ;
     return x>?:-;
 }

 bool operator==(const Point &a,const Point &b){
     return dcmp(a.x-b.x) ==  && dcmp(a.y-b.y) == ;
 }

 Vector operator-(Point a,Point 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 b){
     return Vector(a.x*b,a.y*b);
 }

 Vector operator/(Vector a,double b){
     return Vector(a.x/b,a.y/b);
 }

 bool operator<(const Point &a,const Point &b){
     return a.x<b.x||(a.x==b.x&&a.y<b.y);
 }

 double Dot(Vector a,Vector b){
     return a.x*b.x + a.y*b.y;
 }

 double Cross(Vector a,Vector b){
     return a.x*b.y - a.y*b.x;
 }

 double Length(Vector a){
     return sqrt(Dot(a,a));
 }

 double Angle(Vector a,Vector b){
     return acos(Dot(a,b) / Length(a) / Length(b));
 }

 Point getLineIntersection(Point a,Vector va , Point b , Vector vb){
     Vector c = a-b;
     double t = Cross(vb,c) / Cross(va,vb);
     return a+va*t;
 }

 bool onSegment(Point a,Point st,Point la){
     return dcmp(Cross(st-a,la-a)) ==  && dcmp(Dot(st-a,la-a)) < ;
 }

 bool onIntersection(Point a , Point b , Point c , Point d){
     double t1 = dcmp(Cross(b-a , c-a)) , t2 = dcmp(Cross(b-a , d-a));
     double t3 = dcmp(Cross(d-c , b-c)) , t4 = dcmp(Cross(d-c , a-c));
     return t1*t2 <  && t3*t4<;
 }

 int main()
 {
   // freopen("test.in","rb",stdin);
     int kase = ;
     int n,x,y,e,c;
     while(~scanf("%d",&n)){
         if(n==) break;

         vp.clear();
         for(int i=;i<n;i++){
             scanf("%d%d",&x,&y);
             po[i].x = x,po[i].y = y;
             vp.push_back(po[i]);
         }
         n--;
         e=n;
         for(int i=;i<n;i++){
             for(int j=i+;j<n;j++){
                 //if(i==j) continue;
                 if(onIntersection(po[i],po[i+],po[j],po[j+])){
                     vp.push_back(getLineIntersection(po[i],po[i+]-po[i],po[j],po[j+]-po[j]));
                 }
             }
         }
         sort(vp.begin(),vp.end());
         c=unique(vp.begin(),vp.end()) - vp.begin();
         for(int i=;i<c;i++){
             for(int j=;j<n;j++){
                 if(onSegment(vp[i],po[j],po[j+]))
                     e++;
             }
         }
         printf("Case %d: There are %d pieces.\n",++kase,e-c+);
     }
     return ;
 }