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

---

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

---

/*************************************************************************
> 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 hi<=eps && jk<=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;
}