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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12161 | Accepted: 3847 |
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!
/*
poj 3304 直线与线段相交
给你n条线段,确定是否存在一条直线，它们的投影到上面时有公共点
如果存在一个这样的直线,那么说明所有的线段能和一条直线相交
对这条直线进行一定的旋转,必定与所有直线的端点中的至少两个相交
所以枚举所有的端点进行判断即可
//注意判断相同点
hhh-2016-05-04 20:48:26
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <cmath>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef long long ll;
const int maxn = 40010;
double eps = 1e-8;
int tot;
int n,m;
double x1,x2,y1,y2;
int sgn(double x)
{
if(fabs(x) < eps) return 0;
if(x < 0)
return -1;
else
return 1;
}
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;
}
};
struct Line
{
Point s,t;
Line() {}
Line(Point _s,Point _t)
{
s = _s;
t = _t;
}
};
int tans[maxn];
Line line[maxn];
Point po[maxn];
Point p;
bool seg_inter_line(Line l1,Line l2)
{
return sgn((l2.s-l1.t) ^ (l1.s-l1.t))*sgn((l2.t-l1.t)^(l1.s-l1.t)) <= 0;
}
bool cal(Line tl)
{
for(int i = 0;i < n;i++)
{
if(!seg_inter_line(tl,line[i]))
{
return false;
}
}
return true;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int flag = 0;
tot = 0;
if(n < 3)
flag = 1;
for(int i = 0; i < n; i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
po[tot++] = Point(x1,y1);
po[tot++] = Point(x2,y2);
line[i] = Line(po[tot-1],po[tot-2]);
}
for(int i = 0; i < tot && !flag; i++)
{
for(int j = i+1; !flag && j < tot; j++)
{
if(po[i].x == po[j].x && po[i].y == po[j].y)
continue;
if(cal(Line(po[i],po[j])))
{
flag = 1;
break;
}
}
}
if(flag)
printf("Yes!\n");
else
printf("No!\n");
}
return 0;
}
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原文地址：http://www.cnblogs.com/Przz/p/5510310.html
