poj 3304 Segments（计算几何基础）

Segments

Description

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x₁ y₁ x₂ y₂ follow, in which (x₁, y₁) and (x₂, y₂) are the coordinates of the two endpoints for one of the segments.

Output

For each test case, your program must output "Yes!", if a line with desired property exists and must output "No!" otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10^-8.

Sample Input

3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0

Sample Output

Yes!
Yes!
No!

Source

Amirkabir University of Technology Local Contest 2006

【题意】

　　给出n条线段两个端点的坐标，问所有线段投影到一条直线上，如果这些所有投影至少相交于一点就输出Yes！，否则输出No！。
【思路】

　　如果有存在这样的直线，过投影相交区域作直线的垂线，该垂线必定与每条线段相交，问题转化为问是否存在一条线和所有线段相交
　　若存在一条直线与所有线段相机相交，此时该直线必定经过这些线段的某两个端点，所以枚举任意两个端点即可

【代码】

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<cstring>
 4 #define FOR(a,b,c) for(int a=(b);a<(c);a++)
 5 using namespace std;
 6
 7 const int N = 100+10;
 8 const double eps = 1e-8;
 9
10 int dcmp(double x) {
11     if(x<eps) return 0; else return x<0? -1:1;
12 }
13
14 struct Line {
15     double x1,y1,x2,y2;
16 }L[N];
17
18 int n;
19
20 double cross(double x1,double y1,double x2,double y2,double x,double y) {
21     return (x2-x1)*(y-y1)-(x-x1)*(y2-y1);
22 }
23 bool judge(double x1,double y1,double x2,double y2) {
24     if(!dcmp(x2-x1) && !dcmp(y2-y1)) return 0;
25     FOR(i,0,n) {
26         if(cross(x1,y1,x2,y2,L[i].x1,L[i].y1)*
27             cross(x1,y1,x2,y2,L[i].x2,L[i].y2)>eps) return 0;
28     }
29     return 1;
30 }
31
32 int main() {
33     int T;
34     scanf("%d",&T);
35     while(T--) {
36         scanf("%d",&n);
37         FOR(i,0,n)
38             scanf("%lf%lf%lf%lf",&L[i].x1,&L[i].y1,&L[i].x2,&L[i].y2);
39         if(n==1) { puts("Yes!"); continue; }
40         bool ans=0;
41         FOR(i,0,n) {
42             FOR(j,i+1,n) {
43                 if(judge(L[i].x1,L[i].y1,L[j].x1,L[j].y1)) ans=1;
44                 if(judge(L[i].x1,L[i].y1,L[j].x2,L[j].y2)) ans=1;
45                 if(judge(L[i].x2,L[i].y2,L[j].x1,L[j].y1)) ans=1;
46                 if(judge(L[i].x2,L[i].y2,L[j].x2,L[j].y2)) ans=1;
47                 if(ans) break;
48             }
49             if(ans) break;
50         }
51         if(ans) puts("Yes!");
52         else puts("No!");
53     }
54     return 0;
55 }