poj3304 Segments

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9898		Accepted: 3050

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!
题意：给出N条线段，问是否存在每条线段的投影交于一点的情况。
也就是说有一条直线经过所有的线段。
题解：要经过这N条线段，通过平移，满足条件，这条直线肯定经过这些线段的至少某两个端点。所以枚举这2*N个端点，看是否这一条直线与这N条线段相交。判断直线与线段相交，可以把线段的两个端点跟直线的位置进行叉积。分别在直线两边(叉积为负)或者某个端点在直线上（叉积为零)则线段与直线相交。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#include <ctime>
#define LL __int64
#define eps 1e-8
using namespace std;
struct point
{
	double x,y;
}l[200],r[200],p[210];
double across(point a,point b,point c)
{
	return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}
int go(point a,point b,point c,point d)
{
	double ans=across(a,b,c)*across(a,b,d);
	if (ans<0 || fabs(ans)<eps) return 1;
	else return 0;
}
int main()
{
	int T,i,j,k,n;
	scanf("%d",&T);
	while (T--){
		scanf("%d",&n);
		int cnt=0;
		for (i=1;i<=n;i++){
			scanf("%lf%lf%lf%lf",&l[i].x,&l[i].y,&r[i].x,&r[i].y);
			p[cnt].x=l[i].x;p[cnt++].y=l[i].y;
			p[cnt].x=r[i].x;p[cnt++].y=r[i].y;
		}
		int flag=1;
		for (i=0;i<cnt && flag;i++){
			for (j=i+1;j<cnt && flag;j++){
				if (p[i].x==p[j].x && p[i].y==p[j].y) continue;
				int ff=1;
				for (k=1;k<=n;k++){
					if (!go(p[i],p[j],l[k],r[k])){
						ff=0;
						break;
					}
				}
				if (ff) flag=0;
			}
		}
		if (flag) cout<<"No!"<<endl;
		else cout<<"Yes!"<<endl;
	}
	return 0;
}