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=0,double y=0):x(x),y(y){}
}po[100005];
typedef Point Vector;

vector<Point> vp;

int dcmp(double x)
{
    if(abs(x)<eps) return 0;
    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;
}

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)) == 0 && dcmp(Dot(st-a,la-a)) < 0;
}

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 < 0 && t3*t4<0;
}

int main()
{
  // freopen("test.in","rb",stdin);
    int kase = 0;
    int n,x,y,e,c;
    while(~scanf("%d",&n)){
        if(n==0) break;

        vp.clear();
        for(int i=0;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=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();
        for(int i=0;i<c;i++){
            for(int j=0;j<n;j++){
                if(onSegment(vp[i],po[j],po[j+1]))
                    e++;
            }
        }
        printf("Case %d: There are %d pieces.
",++kase,e-c+2);
    }
    return 0;
}