【poj-1066】判断线段相交

题目链接：https://vjudge.net/problem/POJ-1066

题意：给出一些直线，直线与直线切割的线段是围墙，只能从围墙中间穿过，问最少穿过几层墙才能到达终点

看了hzwer的，觉得好有道理，虽然说是中点，但其实直接求每个端点就ok了

 1 /*************************************************************************
 2     > File Name: poj1066.cpp
 3 # File Name: poj1066.cpp
 4 # Author : xiaobuxie
 5 # QQ : 760427180
 6 # Email:[email protected]
 7 # Created Time: 2019年09月21日 星期六 15时11分27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 pq priority_queue<int,vector<int>,greater<int> >
23 const double eps = 1e-8;
24 const int N=50;
25 int sgn(double x){ 
26     if(fabs(x)<eps) return 0;
27     if(x<0) return -1;
28     return 1;
29 }
30 struct Point{
31     double x,y;
32     Point operator - (const Point& b)const{
33         return (Point){x-b.x,y-b.y};
34     }
35     double operator ^ (const Point& b)const{
36         return x*b.y-b.x*y;
37     }
38 }P[N<<2];
39 struct Line{
40     Point s,e;
41 }L[N];
42 bool inter(Line l1,Line l2){
43     return 
44         max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
45         max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
46         max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
47         max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
48         sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s)) <= 0 &&
49         sgn((l1.s-l2.s)^(l2.e-l1.s))*sgn((l1.e-l2.s)^(l2.e-l2.s)) <= 0;
50 }
51 int main(){
52     int n; scanf("%d",&n);
53     if(!n){
54         printf("Number of doors = 1");
55         return 0;
56     }
57     double x1,x2,y1,y2,ex,ey;
58     for(int i=1;i<=n;++i){
59         scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
60         L[i]=(Line){(Point){x1,y1},(Point){x2,y2}};
61         P[i*2]=(Point){x1,y1}; P[i*2-1]=(Point){x2,y2};
62     }
63     scanf("%lf %lf",&ex,&ey);
64     Point ed=(Point){ex,ey};
65     int ans=10000000;
66     for(int i=1;i<=2*n;++i){
67         Line l1=(Line){ed,P[i]};
68         int tem=0;
69         for(int j=1;j<=n;++j){
70             if(i==2*j || i==2*j-1) continue;
71             if(inter(l1,L[j])) ++tem;
72         }
73         ans=min(ans,tem);
74         //cerr<<ans<<endl;
75     }
76     //cerr<<ans<<endl;
77     printf("Number of doors = %d",ans+1);
78     return 0;
79 }

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