poj3304 Segments

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15967		Accepted: 5063

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

大致题意：给一些线段，问有没有一条直线，使得这些线段的投影在直线上有公共点.

分析：投影实际上就是做垂线,如果一条直线穿过所有线段，做这条直线的垂线，垂足就是公共点.这是问题的第一步转化.那么这条直线怎么求呢？两点确定一条直线，这个点有无穷多个，不太好确定.事实上这两个点可以由线段的任意两个端点确定.为什么呢.假设一条直线穿过了所有的线段，每条线段就是阻碍点.直线可以在线段上自由挪动，旋转，直到碰到阻碍点.感性地理解一下，如果一条直线穿过两个端点并且穿过所有线段，那么它就是满足条件的，不会有这种情况：没有穿过两个端点的直线穿过所有线段，却有穿过所有线段的直线.那么就很好做了，枚举两个端点，利用叉积判断.

           坑点：n = 1特判掉.

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int T,n;
bool flag = false;
const double eps = 1e-8;

struct node
{
    double x,y;
};

struct node2
{
    node p1,p2;
}e[110];

node sub(node a,node b)
{
    node temp;
    temp.x = a.x - b.x;
    temp.y = a.y - b.y;
    return temp;
}

double det(node a,node b)
{
    return a.x * b.y - a.y * b.x;
}

void check(node a,node b)
{
    if (sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)) < eps)
        return;
    for (int i = 1; i <= n; i++)
    {
        double temp1 = det(sub(b,a),sub(e[i].p1,a));
        double temp2 = det(sub(b,a),sub(e[i].p2,a));
        if (temp1 * temp2 > eps)
            return;
    }
    flag = true;
}

int main()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);
        flag = false;
        for (int i = 1; i <= n; i++)
            scanf("%lf%lf%lf%lf",&e[i].p1.x,&e[i].p1.y,&e[i].p2.x,&e[i].p2.y);
        if (n <= 2)
        {
            puts("Yes!");
            continue;
        }
        for (int i = 1; i <= n; i++)
        {
            for (int j = i + 1; j <= n; j++)
            {
                check(e[i].p1,e[j].p1);
                check(e[i].p1,e[j].p2);
                check(e[i].p2,e[j].p1);
                check(e[i].p2,e[j].p2);
                if (flag)
                    break;
            }
            if (flag)
                break;
        }
        if (flag)
            puts("Yes!");
        else
            puts("No!");
    }

    return 0;
}