HDU 1086You can Solve a Geometry Problem too(判断线段相交模板题)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<sstream>
#include<fstream>
#include<vector>
#include<map>
#include<stack>
#include<list>
#include<set>
#include<queue>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1 | 1
using namespace std;
const int maxn=100005,inf=1<<29;
int dir[][2]={ {0,1},{-1,0},{0,-1},{1,0},{-1,1},{-1,-1},{1,-1},{1,1}};
int n,m,t;
double EPS=1e-10;
double add(double a,double b)
{
    if(abs(a+b)<EPS*(abs(a)+abs(b))) return 0;
    return a+b;
}
struct P
{
    double x,y;
    P(){}
    P(double x,double y):x(x),y(y){}
    P operator +(P p)
    {
        return P(add(x,p.x),add(y,p.y));
    }
    P operator -(P p)
    {
        return P(add(x,-p.x),add(y,-p.y));
    }
    P operator *(double d)
    {
        return P(x*d,y*d);
    }
    double dot(P p)//内积
    {
        return add(x*p.x,y*p.y);
    }
    double det(P p)//外积
    {
        return add(x*p.y,-p.x*y);
    }
};
struct v
{
    P p,q;
}a[maxn];
bool on_seg(P p1,P p2,P q)//判断点q是否在线段p1p2上
{
    return (p1-q).det(p2-q)==0&&(p1-q).dot(p2-q)<=0;
}
P intersection(P p1,P p2,P q1,P q2)//计算两线段的交点
{
    return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));
}
bool judge(v s1,v s2)
{
    if((s1.p-s1.q).det(s2.p-s2.q)==0)
    {
        return on_seg(s1.p,s1.q,s2.p)||
                    on_seg(s1.p,s1.q,s2.q)||
                    on_seg(s2.p,s2.q,s1.p)||
                    on_seg(s2.p,s2.q,s1.q);
    }
    else
    {
        P r=intersection(s1.p,s1.q,s2.p,s2.q);
        //cout<<"x = "<<r.x<<" y= "<<r.y<<endl;
        return on_seg(s1.p,s1.q,r)&&on_seg(s2.p,s2.q,r);
    }
}
int main()
{

    while(cin>>n,n)
    {
        for(int i=0;i<n;i++) cin>>a[i].p.x>>a[i].p.y>>a[i].q.x>>a[i].q.y;
        //cin>>s2.p.x>>s2.p.y>>s2.q.x>>s2.q.y;
        int cnt=0;
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
            if(judge(a[i],a[j])) cnt++;
        cout<<cnt<<endl;
    }
    return 0;
}