POJ3304 Segments

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5604		Accepted: 1630

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

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

const double eps = 1e-8;

struct LPoint{
	double x, y;
};

struct LLine{
	LPoint bg, ed;
}line[110];

int dblcmp(double a, double b){
	if (a - b >= eps) return 1;
	if (b - a >= eps) return -1;
	return 0;
}

double VecPro(double x1, double y1, double x2, double y2){
	return (x1 * y2 - x2 * y1);
}

double NumPro(double x1, double y1, double x2, double y2){
	return (x1 * x1 + y1 * y2);
}
//直线ab，与线段pq是否相交
int IsCross(LPoint a, LPoint b, LPoint p, LPoint q){
	if (dblcmp(VecPro(a.x - b.x, a.y - b.y, p.x - b.x, p.y - b.y), 0) *
	    dblcmp(VecPro(a.x - b.x, a.y - b.y, q.x - b.x, q.y - b.y), 0) <= 0)
		return 1;
	else return 0;
}

double Dis(LPoint a, LPoint b){
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

int main(){
	int i, j, k;
	int n, T;
	scanf("%d", &T);
	while(T--){
		scanf("%d", &n);
		for (i = 0; i < n; i++){
			scanf("%lf %lf %lf %lf", &line[i].bg.x, &line[i].bg.y, &line[i].ed.x, &line[i].ed.y);
		}
		if (n <= 2) k = n;
		else{
			k = 0;
			for (i = 0; i < n; i++){
				for (j = i + 1; j < n; j++){
					if (dblcmp(Dis(line[i].bg, line[j].bg), 0) > 0){
						for (k = 0; k < n; k++){
							if (!IsCross(line[i].bg, line[j].bg, line[k].bg, line[k].ed)) break;
						}
						if (k >= n) break;
					}
					if (dblcmp(Dis(line[i].bg, line[j].ed), 0) > 0){
						for (k = 0; k < n; k++){
							if (!IsCross(line[i].bg, line[j].ed, line[k].bg, line[k].ed)) break;
						}
						if (k >= n) break;
					}
					if (dblcmp(Dis(line[i].ed, line[j].bg), 0) > 0){
						for (k = 0; k < n; k++){
							if (!IsCross(line[i].ed, line[j].bg, line[k].bg, line[k].ed)) break;
						}
						if (k >= n) break;
					}
					if (dblcmp(Dis(line[i].ed, line[j].ed), 0) > 0){
						for (k = 0; k < n; k++){
							if (!IsCross(line[i].ed, line[j].ed, line[k].bg, line[k].ed)) break;
						}
						if (k >= n) break;
					}
				}
				if (k >= n) break;
			}
		}
		if (k >= n) printf("Yes!\n");
		else printf("No!\n");
	}	
	return 0;
}
/*
题意：给n个线段，问是否存在一直线，使所有线段在该直线上投影相交于同一点
转帖discuss里hanjialong的思路：
	首先题中的要求等价于：存在一条直线l和所有的线段都相交。
	证明：若存在一条直线l和所有线段相交，作一条直线m和l垂直，则m就是题中要求的直线，所有线段投影的一个公共点即为垂足。（l和每条线段的交点沿l投影到m上的垂足处）
	反过来，若存在m，所有线段在m上的投影有公共点，则过这点垂直于m作直线l，l一定和所有线段相交。

	然后证存在l和所有线段相交等价于存在l过某两条线段的各一个端点且和所有线段相交。
	充分性显然。必要性：若有l和所有线段相交，则可保持l和所有线段相交，左右平移l到和某一线段交于端点停止（“移不动了”）。然后绕这个交点旋转。也是转到“转不动了”（和另一线段交于其一个端点）为止。这样就找到了一个新的l。

	于是本题可归结为枚举两两线段的各一个端点，连一条直线，再判断剩下的线段是否都和这条直线有交点。
计算几何还是难想啊...
然后就是俩点注意，n <= 2特判为yes, 作为直线的俩点不能重合
dblcmp写错了，贡献一WA...真是不细心啊...
*/