【计算几何】 POJ 1127 Jack Straws 判断线段是否相交

给出n个线段

相交有传递关系

（当个模板）

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#define cler(arr, val)    memset(arr, val, sizeof(arr))
#define FOR(i,a,b)  for(int i=a;i<=b;i++)
#define IN   freopen ("in.txt" , "r" , stdin);
#define OUT  freopen ("out.txt" , "w" , stdout);
typedef long long  LL;
const int MAXN = 10+5;
const int MAXM = 550000;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
const double eps= 1e-8;
double add(double a,double b)//去除误差
{
    if(abs(a+b)<eps*(abs(a)+abs(b)))
        return 0;
    else return a+b;
}
struct point
{
    double x,y;
    point(){}
    point(double x,double y): x(x),y(y){}
    point operator + (point p)
    {
        return point(add(x,p.x),add(y, p.y));
    }
    point operator - (point p)
    {
        return point(add(x,-p.x),add(y,-p.y));
    }
    point operator * (double d)
    {
        return point(x*d,y*d);
    }
    double dot(point p)//点积
    {
        return  add(x*p.x ,y*p.y);
    }
    double det(point p)//叉积
    {
        return  add(x*p.y ,-y*p.x);
    }
};
bool onseg(point p1,point p2,point q)
{
    return (p1-q).det(p2-q)==0 && (p1-q).dot(p2-q) <=0;
}
point intersection(point p1,point p2,point q1,point q2)
{
    return p1 + (p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));
}
point p[MAXN],q[MAXN];
bool g[MAXN][MAXN];
void solve(int n)
{
    for(int i=1;i<=n;i++)
    {
        g[i][i]=true;
        for(int j=1;j<i;j++)
        {
            if((p[i]-q[i]).det((p[j]-q[j]))==0)//平行时
            {
                g[i][j]=g[j][i]=onseg(p[i],q[i],p[j])
                                |onseg(p[i],q[i],q[j])
                                |onseg(p[j],q[j],p[i])
                                |onseg(p[j],q[j],q[i]);
            }
            else
            {
                point r =intersection(p[i],q[i],p[j],q[j]);
                g[i][j]=g[j][i]=onseg(p[i],q[i],r)&&onseg(p[j],q[j],r);
            }
        }
    }
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                g[i][j]|=g[i][k]&&g[k][j];
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif

    int n,a,b;
    while(scanf("%d",&n),n)
    {
        for(int i=1;i<=n;i++)
            cin>>p[i].x>>p[i].y>>q[i].x>>q[i].y;
        cler(g,false);
        solve(n);
        while(scanf("%d%d",&a,&b),a+b)
        {
            g[a][b]?cout<<"CONNECTED"<<endl:cout<<"NOT CONNECTED"<<endl;
        }
    }
    return 0;
}