HDU 1086 You can Solve a Geometry Problem too（ 判断线段是否相交 水题 ）

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链接：传送门

题意：给出 n 个线段找到交点个数

思路：数据量小，直接暴力判断所有线段是否相交

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/*************************************************************************
    > File Name: hdu1086.cpp
    > Author:    WArobot 
    > Blog:      http://www.cnblogs.com/WArobot/ 
    > Created Time: 2017年05月07日 星期日 23时34分32秒
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;

#define eps 1e-10
struct point{ double x,y; };
struct V{ point s,e; };

bool inter(point a,point b,point c,point d){
	if( min(a.x,b.x) > max(c.x,d.x) ||
		min(a.y,b.y) > max(c.y,d.y) ||
		min(c.x,d.x) > max(a.x,b.x) ||
		min(c.y,d.y) > max(a.y,b.y)
	)return 0;
	double h,i,j,k;
	h = (b.x-a.x)*(c.y-a.y) - (b.y-a.y)*(c.x-a.x);
	i = (b.x-a.x)*(d.y-a.y) - (b.y-a.y)*(d.x-a.x);
	j = (d.x-c.x)*(a.y-c.y) - (d.y-c.y)*(a.x-c.x);
	k = (d.x-c.x)*(b.y-c.y) - (d.y-c.y)*(b.x-c.x);
	return h*i<=eps && j*k<=eps;
}
int main(){
	int n;
	V vct[110];
	while(~scanf("%d",&n) && n){
		for(int i=0;i<n;i++)	scanf("%lf%lf%lf%lf",&vct[i].s.x,&vct[i].s.y,&vct[i].e.x,&vct[i].e.y);
		int cnt = 0;
		for(int i=0;i<n;i++){
			for(int j=i+1;j<n;j++){
				if( inter(vct[i].s,vct[i].e,vct[j].s,vct[j].e) )	cnt++;
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
}