poj3304计算几何直线与线段关系

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!
一开始看不懂题意只好搜题意，看懂了题意之后还是花了两个多小时wa了七遍，只好看题解，可能是精度的地方写搓了>.<坑爹啊
叉积判断线段的两端点是不是在直线的两侧

#include<map>
#include<set>
#include<list>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007

using namespace std;

const double eps=1e-;
const int N=,maxn=,inf=0x3f3f3f3f;

struct point{
    double x,y;
};
struct line{
   point a,b;
}l[N];

int n;
double mul(point p,point u,point v)
{
    return (u.x-v.x)*(p.y-u.y)-(u.y-v.y)*(p.x-u.x);
}
bool ok(point u,point v)
{
    if(fabs(u.x-v.x)<eps&&fabs(u.y-v.y)<eps)return ;
    for(int i=;i<n;i++)
        if(mul(l[i].a,u,v)*mul(l[i].b,u,v)>=eps)
           return ;
    return ;
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        cin>>n;
        for(int i=;i<n;i++)
            cin>>l[i].a.x>>l[i].a.y>>l[i].b.x>>l[i].b.y;
        if(n<=)
        {
            cout<<"Yes!"<<endl;
            continue;
        }
        bool flag=;
        for(int i=;i<n&&!flag;i++)
        {
            if(ok(l[i].a,l[i].b))flag=;
            for(int j=i+;j<n&&!flag;j++)
                if(ok(l[i].a,l[j].a)||ok(l[i].a,l[j].b)||ok(l[i].b,l[j].a)||ok(l[i].b,l[j].b))
                   flag=;
        }
        if(flag)cout<<"Yes!"<<endl;
        else cout<<"No!"<<endl;
    }
    return ;
}