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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 x1y1x2y2 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!
# include <iostream> # include <cmath> # include <cstdio> using namespace std; # define EPS 1e-8 int n; struct Segment { double x1,x2,y1,y2; }segment[1000]; double distance(double x1,double y1,double x2,double y2) { return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); } double fun(double x1,double y1,double x2,double y2,double x0,double y0) { return (x2-x1)*(y0-y1)-(x0-x1)*(y2-y1); } bool search(double x1,double y1,double x2,double y2) { int i; //推断是不是同一条直线; if(distance(x1,y1,x2,y2)<EPS) return false; //利用叉积定理。推断是不是和其它的线段相交： for(i=0;i<n;i++) { if(fun(x1,y1,x2,y2,segment[i].x1,segment[i].y1)* fun(x1,y1,x2,y2,segment[i].x2,segment[i].y2)>EPS) return false; } return true; } int main () { //freopen("a.txt","r",stdin); int t; scanf("%d",&t); while(t--) { scanf("%d",&n); int i; for(i=0;i<n;i++) scanf("%lf%lf%lf%lf",&segment[i].x1,&segment[i].y1,&segment[i].x2,&segment[i].y2); if(n==1) { printf("Yes!\n"); continue;} int ans=0; //将全部的线段都列举出来。 for(i=0;i<n;i++) for(int j=i+1;j<n;j++) { if(search(segment[i].x1,segment[i].y1,segment[j].x1,segment[j].y1)|| search(segment[i].x1,segment[i].y1,segment[j].x2,segment[j].y2)|| search(segment[i].x2,segment[i].y2,segment[j].x1,segment[j].y1)|| search(segment[i].x2,segment[i].y2,segment[j].x2,segment[j].y2)) ans=1; } if(ans) printf("Yes!\n"); else printf("No!\n"); } return 0; }
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