POJ 3304 Segments（判断直线和线段相交）

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12150		Accepted: 3845

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

原博：http://blog.csdn.net/wangjian8006

题目大意：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！。
解题思路：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。
这里要主要的地方就是，题目说如果两个点的距离小于1e-8就等价于一点，所以要考虑重点

/*
Memory 200K
Time   32MS 
*/
#include <iostream>
#include <math.h>
using namespace std;
#define MAXM 110
#define EPS 1e-8

typedef struct{
	double x1,y1,x2,y2;
}Segment;

Segment segment[MAXM];
int n;

double distance(double x1,double y1,double x2,double y2){
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

double corss(double x1,double y1,double x2,double y2,double x,double y){
	return (x2-x1)*(y-y1)-(x-x1)*(y2-y1);
}

bool judge(double x1,double y1,double x2,double y2){
	int i;
	if(distance(x1,y1,x2,y2)<EPS) return 0;
	for(i=0;i<n;i++){
		if(corss(x1,y1,x2,y2,segment[i].x1,segment[i].y1)*
			corss(x1,y1,x2,y2,segment[i].x2,segment[i].y2)>EPS) return 0;
	}
	return 1;
}

int main(){
	int t,i,j,ans;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		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!
");continue;}

		ans=0;
		for(i=0;i<n && !ans;i++)
			for(j=i+1;j<n && !ans;j++){
				if(judge(segment[i].x1,segment[i].y1,segment[j].x1,segment[j].y1) ||
					judge(segment[i].x1,segment[i].y1,segment[j].x2,segment[j].y2) ||
					judge(segment[i].x2,segment[i].y2,segment[j].x1,segment[j].y1) ||
					judge(segment[i].x2,segment[i].y2,segment[j].x2,segment[j].y2))
					ans=1;
			}
		if(ans) printf("Yes!
");
		else printf("No!
");
	}
	return 0;