[POJ 3304]Segments

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14918		Accepted: 4728

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！。

题解：

首先转化题意，把问题转化为是否存在一条直线与每条线段都有交点。

然后两两枚举线段，每对线段可以产生四条直线，再分别判断是否可以经过所有线段即可。

以上题意转化证明：大佬的博客

本题需要计算几何的相关知识，具体来说应该是跨立实验判断直线与线段有无交点。

特别注意：

1.精度问题，小于精度就直接判断为相等，特别地，两个点距离小于精度的判断为同一个点，此时不能确定一条直线。

2.所得直线有可能和某一条线段重合，所以不能加i!=j的判断。

以下是AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstring>
#include<cstdlib>
#define exp (1e-8)
using namespace std;
int t,n;
struct Vector
{
    double x,y;
    Vector(){}
    Vector(int xx,int yy){x=xx;y=yy;}
    Vector operator+(Vector b)
    {
        Vector ans;
        ans.x=x+b.x;ans.y=y+b.y;
        return ans;
    }
    Vector operator-(Vector b)
    {
        Vector ans;
        ans.x=x-b.x;ans.y=y-b.y;
        return ans;
    }
    double operator*(Vector b)
    {
        return x*b.y-b.x*y;
    }
    double operator^(Vector b)
    {
        return x*b.x+y*b.y;
    }
};
struct node
{
    Vector a,b;
}a[101];
bool find(Vector s1,Vector s2)
{
    if(fabs(s1.x-s2.x)<exp&&fabs(s1.y-s2.y)<exp)return false;
    int i,cnt=0;
    for(i=1;i<=n;i++)
    {
        if(((s1-s2)*(a[i].a-s2))*((s1-s2)*(a[i].b-s2))>exp)return false;
    }
    return true;
}
bool check(int x,int y)
{
    bool b=0;
    b|=find(a[x].a,a[y].a);
    b|=find(a[x].a,a[y].b);
    b|=find(a[x].b,a[y].a);
    b|=find(a[x].b,a[y].b);
    return b;
}
int main()
{
    int i,j;
    bool ok=0;
    scanf("%d",&t);
    while(t--)
    {
        ok=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)scanf("%lf%lf%lf%lf",&a[i].a.x,&a[i].a.y,&a[i].b.x,&a[i].b.y);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(check(i,j))ok=1;
            }
        }
        if(ok)printf("Yes!
");
        else printf("No!
");
    }
}