| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7739 | Accepted: 2316 |
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!
Source
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
/************************************************************
* Author : kuangbin
* Email : kuangbin2009@126.com
* Last modified : 2013-07-13 20:57
* Filename : POJ3304Segments.cpp
* Description :
* *********************************************************/ #include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h> using namespace std; const double eps = 1e-;
int sgn(double x)
{
if(fabs(x) < eps)return ;
if(x < ) return -;
return ;
}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y)
{
x = _x;y = _y;
}
Point operator -(const Point &b)const
{
return Point(x - b.x,y - b.y);
}
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
};
double xmult(Point p0,Point p1,Point p2) //p0p1 X p0p2
{
return (p1-p0)^(p2-p0);
}
bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交
{
return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e)) <= ;
}
double dist(Point a,Point b)
{
return sqrt( (b - a)*(b - a) );
}
const int MAXN = ;
Line line[MAXN];
bool check(Line l1,int n)
{
if(sgn(dist(l1.s,l1.e)) == )return false;
for(int i = ;i < n;i++)
if(Seg_inter_line(l1,line[i]) == false)
return false;
return true;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
double x1,y1,x2,y2;
for(int i = ; i < n;i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
line[i] = Line(Point(x1,y1),Point(x2,y2));
}
bool flag = false;
for(int i = ;i < n;i++)
for(int j = ; j < n;j++)
if(check(Line(line[i].s,line[j].s),n) || check(Line(line[i].s,line[j].e),n)
|| check(Line(line[i].e,line[j].s),n) || check(Line(line[i].e,line[j].e),n) )
{
flag = true;
break;
}
if(flag)
printf("Yes!\n");
else printf("No!\n");
}
return ;
}