POJ 1127 Jack Straws 几何基础

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <cmath>
 5 using namespace std;
 6 const int N = 20;
 7 const int M = 110;
 8 double EPS = 1e-10;
 9 double add(double a,double b) {
10     if(abs(a+b) < EPS * (abs(a) + abs(b))) return 0;
11     return a + b;
12 }
13 struct Point{
14     double x, y;
15     Point(){}
16     Point(double x, double y) :x(x), y(y) {}
17     Point operator + (Point p) {
18         return Point(add(x, p.x), add(y, p.y));
19     }
20     Point operator - (Point p) {
21         return Point(add(x, -p.x), add(y, -p.y));
22     }
23     Point operator * (double d) {
24         return Point(x*d, y*d);
25     }
26     double dot(Point p) {    //内积
27         return add(x * p.x, y * p.y);
28     }
29     double det(Point p) {    //外积
30         return add(x * p.y, -y * p.x);
31     }
32 };
33 
34 bool on_seg(Point p1, Point p2, Point q) {      //判断点q是否在线段p1-p2上
35     return (p1 - q).det(p2 - q) == 0 && (p1 - q).dot(p2 - q) <= 0;
36 }
37 Point intersection(Point p1, Point p2, Point q1, Point q2) {    //计算直线p1-p2与直接q1-q2的交点
38     return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
39 }
40 int n;
41 Point p[N], q[N];
42 int m;
43 bool g[N][N];
44 void solve() {
45     for(int i = 1; i <= n; i ++) {
46         g[i][i] = true;
47         for(int j = 1; j < i; j ++) {
48             if((p[i] - q[i]).det(p[j] - q[j]) == 0) {
49                 g[i][j] = g[j][i] = on_seg(p[i], q[i], q[j])
50                                     || on_seg(p[i], q[i], p[j])
51                                     || on_seg(p[j], q[j], q[i])
52                                     || on_seg(p[j], q[j], p[i]);
53             } else {
54                 Point r = intersection(p[i], q[i], p[j], q[j]);
55                 g[i][j] = g[j][i] = on_seg(p[i], q[i], r) && on_seg(p[j], q[j], r);
56             }
57         }
58     }
59     for(int k = 1; k <= n; k ++) {
60         for(int i = 1; i <= n; i ++) {
61             for(int j = 1; j <= n; j ++) {
62                 g[i][j] |= g[i][k] && g[k][j];
63             }
64         }
65     }
66 }
67 int main() {
68     while(cin >> n && n) {
69         memset(g, false, sizeof(g));
70         for(int i = 1; i <= n; i ++) {
71             cin >> p[i].x >> p[i].y >> q[i].x >> q[i].y;
72         }
73         int a, b;
74         solve();
75         while(cin >> a >> b && (a+b)) {
76             if(g[a][b]) printf("CONNECTED\n");
77             else printf("NOT CONNECTED\n");
78         }
79     }
80     return 0;
81 }