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

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1086

此题是判断线段相交模板题

图片来自于《挑战程序设计竞赛》计算几何那个章节。

下面是AC代码：

#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 | 1using 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;}

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