POJ 3304 Segments

 

Segments

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3479 Accepted: 941

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

321.0 2.0 3.0 4.04.0 5.0 6.0 7.030.0 0.0 0.0 1.00.0 1.0 0.0 2.01.0 1.0 2.0 1.030.0 0.0 0.0 1.00.0 2.0 0.0 3.01.0 1.0 2.0 1.0

Sample Output

Yes!Yes!No!

#include <iostream>#include <cmath>#include <string>#include <cstdio>const int MAXX = 250;const double eps = 1e-8;struct Lpoint{    double x, y;};struct Llineseg{    Lpoint a, b;};Lpoint end[MAXX];Llineseg num[MAXX];double xmulti( Lpoint p1, Lpoint p2, Lpoint p0 ){    return ((p1.x - p0.x ) * (p2.y - p0.y ) - (p2.x - p0.x) * ( p1.y - p0.y ));}int equal( Lpoint a, Lpoint b ){    if( fabs( a.x - b.x ) < eps && fabs( a.y - b.y ) < eps )        return 0;    else        return 1;}bool intersect( Llineseg a, Lpoint b, Lpoint c ){    if( xmulti( a.a, b, c ) * xmulti( a.b, b, c ) <= 0 )        return 1;    else        return 0;}bool run( int n ){    int i, j, k;    for( i = 0; i < n * 2 - 1; i++ )        for( j = i + 1; j < 2 * n; j++ )        if(equal( end[i], end[j] ))        {            for( k = 0; k < n; k++ )            {                if( !intersect( num[k], end[i], end[j] ) )                    break;            }            if( k == n )            {                return true;            }        }    return false;}int main(){    int i, j, k;    double x, y;    int t;    scanf("%d", &t );    while( t-- )    {        int n;        scanf("%d", &n);        for( i = 0, j = 0; i < n; i++, j++ )        {            scanf("%lf%lf", &x, &y);            end[j].x = x;            end[j].y = y;            num[i].a.x = x;            num[i].a.y = y;            j++;            scanf("%lf%lf", &x, &y);            end[j].x = x;            end[j].y = y;            num[i].b.x = x;            num[i].b.y = y;        }        if( run(n) )        {            printf("Yes!/n");        }        else        {            printf("No!/n");        }    }    return 0;}