【模板】 计几判断线段是否相交

题目链接：https://vjudge.net/problem/POJ-2653

参考自kuangbin：https://www.cnblogs.com/kuangbin/p/3189750.html

线段是否相交：

 1 /*************************************************************************
 2     > File Name: poj2653.cpp
 3 # File Name: poj2653.cpp
 4 # Author : xiaobuxie
 5 # QQ : 760427180
 6 # Email:760427180@qq.com
 7 # Created Time: 2019年09月20日 星期五 17时46分50秒
 8  ************************************************************************/
 9 
10 #include<iostream>
11 #include<cstdio>
12 #include<map>
13 #include<cmath>
14 #include<cstring>
15 #include<set>
16 #include<queue>
17 #include<vector>
18 #include<algorithm>
19 using namespace std;
20 typedef long long ll;
21 #define inf 0x3f3f3f3f
22 #define eps 1e-8
23 #define pq priority_queue<int,vector<int>,greater<int> >
24 ll gcd(ll a,ll b){
25     if(a<b) return gcd(b,a);
26     return b==0?a:gcd(b,a%b);
27 }
28 
29 const int N=1e5+9;
30 struct Point{
31     double x,y;
32     Point(){}
33     Point(double _x,double _y){
34         x=_x; y=_y;
35     }
36     Point operator - (const Point& b)const{
37         return Point(x-b.x,y-b.y);
38     }
39     double operator * (const Point& b)const{
40         return x*b.x+y*b.y;
41     }
42     double operator ^ (const Point& b)const{
43         return x*b.y-b.x*y;
44     }
45 };
46 struct Line{
47     Point s,e;
48     Line(){}
49     Line(Point _s,Point _e){
50         s=_s; e=_e;
51     }
52 }L[N];
53 int sgn(double x){
54     if(fabs(x)<eps) return 0;
55     if(x<0) return -1;
56     return 1;
57 }
58 int ans[N];
59 
60 bool insert(Line l1,Line l2){
61     return 
62         max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
63         max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
64         max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
65         max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
66         sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s))<=0 &&
67         sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s))<=0;
68 }
69 int main(){
70     int n;
71     double x1,x2,y1,y2;
72     int cnt=0;
73     while(~scanf("%d",&n) && n){
74         cnt=0;
75         for(int i=1;i<=n;++i){
76             scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
77             L[i]=Line(Point(x1,y1),Point(x2,y2));
78         }
79         for(int i=1;i<=n;++i){
80             bool ok=1;
81             for(int j=i+1;j<=n;++j){
82                 if(insert(L[i],L[j])){
83                     ok=0;
84                     break;
85                 }
86             }
87             if(ok) ans[++cnt]=i;
88         }
89         printf("Top sticks: ");
90         for(int i=1;i<cnt;++i) printf("%d, ",ans[i]);
91         printf("%d.\n",ans[cnt]);
92     }
93     return 0;
94 }

标签：计几,min,线段,l2,&&,l1,include,ll,模板
来源： https://www.cnblogs.com/xiaobuxie/p/11560236.html