POJ_3304_Segments_线段判断是否相交

POJ_3304_Segments_线段判断是否相交

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!

---

在二维空间中给定n个线段，写一个程序，它决定是否存在一条线，这样在投射出这些段之后，所有的投影段至少有一个点是相同的。

假设存在这样的直线，那么过所有的投影的交点做这条直线的垂线一定和所有线段相交，问题转化为是否存在一条直线与所有线段相交。
并且这条线段能通过平移或旋转使它经过某两个线段的端点。于是只需要把所有点存起来，枚举直线。
注意这道题可能有重点。

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef double f2;
#define eps 1e-8
struct Point {
	f2 x,y;
	Point() {}
	Point(f2 x_,f2 y_) :
		x(x_),y(y_) {}
	Point operator + (const Point &p) {return Point(x+p.x,y+p.y);}
	Point operator - (const Point &p) {return Point(x-p.x,y-p.y);}
	Point operator * (f2 rate) {return Point(x*rate,y*rate);}
};
f2 dot(const Point &p1,const Point &p2) {return p1.x*p2.x+p1.y*p2.y;}
f2 cross(const Point &p1,const Point &p2) {return p1.x*p2.y-p1.y*p2.x;}
f2 fabs(f2 x) {return x>0?x:-x;}
int n;
Point a[120],b[120],c[220],p1,p2;
bool judge(f2 x,f2 y) {
	if(fabs(x)<eps||fabs(y)<eps) return 1;
	return (x<-eps&&y>eps)||(x>eps&&y<-eps);
}
bool check() {
	int i;
	Point t=p1-p2;
	for(i=1;i<=n;i++) {
		Point d1=a[i]-p2,d2=b[i]-p2;
		f2 tmp1=cross(t,d1),tmp2=cross(t,d2);
		if(!judge(tmp1,tmp2)) return 0;
	}
	return 1;
}
int main() {
	int T;
	scanf("%d",&T);
	while(T--) {
		int flg=0;
		scanf("%d",&n);
		int i,j;
		for(i=1;i<=n;i++) {
			scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&b[i].x,&b[i].y);
			c[i]=a[i];
			c[i+n]=b[i];
		}
		for(i=1;i<=2*n;i++) {
			for(j=i+1;j<=2*n;j++) {
				if(fabs(c[i].x-c[j].x)>eps||fabs(c[i].y-c[j].y)>eps) {
					p1=c[i]; p2=c[j];
					if(check()) {
						flg=1; break;
					}
				}
			}
			if(flg) break;
		}
		puts(flg?"Yes!":"No!");
	}
}