POJ2284 That Nice Euler Circuit （欧拉公式）（计算几何 线段相交问题）

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#define inf 0xffffff
#include<iostream>
#include<cmath>
#define NUM 22
#include <algorithm>
using namespace std;


const double eps=1e-6;
struct point {
    double x,y;
    point(double a=0,double b=0) {
        x=a;
        y=b;
    }
};
bool operator< (point a, point b) {
    return a.x<b.x||a.x==b.x&&a.y<b.y;
}
bool operator == (point a,point b) {
    return abs(a.x-b.x)<eps&&abs(a.y-b.y)<eps;
}
struct Lineseg {
    point s,e;
    Lineseg(point a, point b) {
        s=a;
        e=b;
    }
};
struct Line {
    double a,b,c;
};
bool online(Lineseg L,point p) { //判断p是否在线段L上
    return abs((L.e.x-L.s.x)*(p.y-L.s.y)-(p.x-L.s.x)*(L.e.y-L.s.y))<eps&&(p.x-L.s.x)*(p.x-L.e.x)<eps&&(p.y-L.s.y)*(p.y-L.e.y)<eps;
}
Line Makeline(Lineseg tmp) { //线段L变成L
    Line L;
    int x1=tmp.s.x;
    int y1=tmp.s.y;
    int x2=tmp.e.x;
    int y2=tmp.e.y;
    if(y2-y1>0) {
        L.a=(y2-y1);
        L.b=(x1-x2);
        L.c=(x2*y1-x1*y2);
    } else {
        L.a=(y1-y2);
        L.b=(x2-x1);
        L.c=(x1*y2-x2*y1);
    }
    return L;
}
bool Lineinter(Line x,Line y,point &q) { //直线X,Y相交于点q
    double d=x.a*y.b-y.a*x.b;
    if(abs(d)<eps)
        return false;
    q.x=(y.c*x.b-x.c*y.b)/d;
    q.y=(y.a*x.c-x.a*y.c)/d;
    return 1;
}


bool Lineseginter(Lineseg aa,Lineseg bb,point &q) { //线段aa，bb如果相交则返回交点q
    Line a,b;
    a=Makeline(aa);
    b=Makeline(bb);
    if(Lineinter(a,b,q))
        return online(aa,q)&&online(bb,q);
    else
        return false;
}
bool cmp(point a ,point b) {
    if(a.x==b.x)
        return a.y<b.y;
    else
        return a.x<b.x;
}
point p[96003];
point inter[98000];
int N;
int main() {
    int m,n;
    int T=0;
    while(scanf("%d",&N),N) {
        m=n=0;
        int cnt=0;
        for(int i=0; i<N; i++)
            scanf("%lf %lf",&p[i].x,&p[i].y);
        for(int i=0; i<N; i++) {
            for(int j=0; j<N; j++) {
                Lineseg L1(p[i],p[(i+1)%N]),L2(p[j],p[(j+1)%N]);
                point q;
                if(Lineseginter(L1,L2,q))
                    inter[cnt++]=q;
            }
        }
        sort(inter,inter+cnt,cmp);
        n=unique(inter,inter+cnt)-inter;//去重复的点
        for(int i=0; i<n; i++) {
            for(int j=0; j<N; j++) {
                Lineseg t(p[j],p[(j+1)%N]);
                if(online(t,inter[i])&&!(t.s==inter[i]))m++;
            }
        }
        T++;
        printf("Case %d: There are %d pieces.\n",T,m+2-n);//欧拉定理
    }
    return 0;
}