POJ 3304(计算几何初步——判断线段和直线相交，加上枚举)

这题有2点要注意：

一丶当n==1时，应特判输出YES

二丶精度控制，当2点的距离相差小于1E-8，近似看作同一点。

View Code

 1 #include<iostream>
 2 #include<cmath>
 3 #define EPS 1e-8
 4 #define MAXN 104
 5 using namespace std;
 6 struct point {
 7     double x, y;    
 8 };
 9 typedef point VECTOR;
10 
11 struct segment {
12     point p[2];    
13 }data[MAXN];
14 
15 bool vis[MAXN];
16 
17 double cross(point &p1, point &p2)
18 {
19     return p2.y*p1.x - p2.x*p1.y;    
20 }
21 
22 bool equal(point &sp, point &ep)
23 {
24     if (sqrt((sp.x - ep.x)*(sp.x - ep.x) + (sp.y - ep.y)*(sp.y - ep.y)) < EPS)return true;
25     return false;    
26 }
27 
28 bool check(point &sp, point &ep, int n)
29 {
30     for (int i(0); i<n; ++i) {
31         if (!vis[i]) {
32             VECTOR fir, sec, thr;
33             fir.x = data[i].p[0].x - sp.x;
34             fir.y = data[i].p[0].y - sp.y;
35             sec.x = ep.x - sp.x;
36             sec.y = ep.y - sp.y;
37             thr.x = data[i].p[1].x - sp.x;
38             thr.y = data[i].p[1].y - sp.y;
39             if (cross(fir, sec)*cross(sec, thr) < -EPS)return false;    
40         }    
41     }
42     return true;    
43 }
44 
45 bool judge(int n)
46 {
47     for (int i(0); i<n; ++i) {
48         vis[i] = true;
49         for (int num = 0; num<2; ++num) {
50             for (int j(i+1); j<n; ++j) {
51                 for (int cnt = 0; cnt<2; ++cnt) {
52                     if (!equal(data[i].p[num],data[j].p[cnt])) {
53                         vis[j] = true;
54                         if (check(data[i].p[num],data[j].p[cnt],n))return true;
55                         vis[j] = false;    
56                     }    
57                 }    
58             }    
59         }
60         vis[i] = false;
61     }
62     return false;            
63 }
64 
65 int main()
66 {
67     int T;
68     scanf("%d",&T);
69     while (T--) {
70         memset(vis,false,sizeof(vis));
71         int n;
72         scanf("%d",&n);
73         for (int i(0); i<n; ++i) {
74             scanf("%lf%lf%lf%lf",&data[i].p[0].x,&data[i].p[0].y,&data[i].p[1].x,&data[i].p[1].y);
75         }
76         if (n == 1) {
77             cout<<"Yes!\n";
78             continue;
79         }
80         if (judge(n)) {
81             cout<<"Yes!\n";    
82         } else {
83             cout<<"No!\n";    
84         }
85     }
86     return 0;    
87 }