POJ 3304 Segments

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7363		Accepted: 2207

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!

Source

Amirkabir University of Technology Local Contest 2006

 

 

 

 

#include<stdio.h>
#include<math.h>

const int eps=1e-8;

struct point{
    double x,y;
}p[410];

double xmult(point a,point b,point c){
    return (a.x-b.x)*(c.y-b.y)-(c.x-b.x)*(a.y-b.y);
}

double abs(double a){
    if(a<0)
        return -a;
    return a;
}

bool work(int n){
    int flag;
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++){
            if(abs(p[j].x-p[i].x)>eps || abs(p[j].y-p[i].y)>eps){
                flag=1;
                for(int k=1;k<=n;k+=2)
                    if(xmult(p[i],p[j],p[k])*xmult(p[i],p[j],p[k+1])>0){
                        flag=0;
                        break;
                    }
                if(flag)
                    return 1;
            }
        }
    return 0;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int cnt=1;
        for(int i=1;i<=n;i++,cnt+=2)
            scanf("%lf%lf%lf%lf",&p[cnt].x,&p[cnt].y,&p[cnt+1].x,&p[cnt+1].y);
        if(work(cnt-1))
            printf("Yes!\n");
        else
            printf("No!\n");
    }
    return 0;
}