poj-3304-Segments-线段与直线相交

判断线段与直线是否相交，注意精度问题。

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<math.h>
using namespace std;
#define eps 0.00000001
#define zero(x) ((fabs(x)<eps)?0:x)
struct segn
{
    double x1,y1;
    double x2,y2;
}seg[110];
int n;
struct line
{
    int leap;
    double k;
    double b;
}ll;
int dos(line a,segn b)
{
    if(a.leap==1)
    {
        if((a.b<=max(b.x1,b.x2)&&a.b>=min(b.x1,b.x2))||(!zero(a.b-b.x1)||!zero(a.b-b.x2)))return 1;
        else return 0;
    }
    double b1,b2,bb;
    b1=b.y1-a.k*b.x1;
    b2=b.y2-a.k*b.x2;
    bb=a.b;
    if((bb<=max(b1,b2)&&bb>=min(b1,b2))||(!zero(bb-b1)||!zero(bb-b2)))return 1;
    return 0;
}
int pan(double x1,double y1,double x2,double y2)
{
    if(zero(x1-x2)==0)
    {
        ll.leap=1;
        ll.b=x1;
    }
    else
    {
        ll.leap=0;
        ll.b=(y1*x2/x1-y2)/(x2/x1-1);
        ll.k=(y1-ll.b)/x1;
    }
    for(int i=1;i<=n;i++)
    {
        if(!dos(ll,seg[i]))return 0;
    }
    return 1;
}
int leap;
int main()
{
    int T,i,j;
    cin>>T;
    while(T--)
    {
        leap=0;
        cin>>n;
        for(i=1; i<=n; i++)
        {
            cin>>seg[i].x1>>seg[i].y1>>seg[i].x2>>seg[i].y2;
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(pan(seg[i].x1,seg[i].y1,seg[i].x2,seg[i].y2))leap=1;
                if(pan(seg[i].x1,seg[i].y1,seg[j].x2,seg[j].y2))leap=1;
                if(pan(seg[i].x1,seg[i].y1,seg[j].x1,seg[j].y1))leap=1;
                if(pan(seg[j].x1,seg[j].y1,seg[i].x2,seg[i].y2))leap=1;
                if(pan(seg[j].x1,seg[j].y1,seg[j].x2,seg[j].y2))leap=1;
                if(pan(seg[j].x2,seg[j].y2,seg[i].x2,seg[i].y2))leap=1;
                if(leap)break;
            }
            if(leap)break;
        }
        if(leap==1)cout<<"Yes!"<<endl;
        else cout<<"No!"<<endl;
    }
    return 0;
}

poj-3304-Segments-线段与直线相交,布布扣,bubuko.com

poj-3304-Segments-线段与直线相交

原文：http://blog.csdn.net/rowanhaoa/article/details/21957001