POJ 3304 segments 线段和直线相交

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14178		Accepted: 4521

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 integern ≤ 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

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 103
#define N 21
#define MOD 1000000
#define INF 1000000009
const double eps = 1e-8;
const double PI = acos(-1.0);
/*
所有线段投射到给定线段上取交集，如果交集距离大于eps 存在！ｓ
*/
int sgn(double x)
{
    if (fabs(x) < eps) return 0;
    if (x < 0) return -1;
    else return 1;
}
struct Point
{
    double x, y;
    Point() {}
    Point(double _x, double _y) :x(_x), y(_y) {}
    Point operator - (const Point& r)const
    {
        return Point(x - r.x, y - r.y);
    }
    double operator ^(const Point& r)const
    {
        return x*r.y - y*r.x;
    }
    double operator * (const Point& r)const
    {
        return x*r.x + y*r.y;
    }
};
double dist(Point a, Point b)
{
    return sqrt((a - b)*(a - b));
}
struct Line
{
    Point s, e;
    Line() {}
    Line(Point _a, Point _B) :s(_a), e(_B) {}
};
vector<Line> v;
bool Seg_inter_line(Line l1, Line l2)
{
    return sgn((l2.s - l1.e) ^ (l1.s - l1.e))*sgn((l2.e - l1.e) ^ (l1.s - l1.e)) <= 0;
}
bool check(Line l)
{
    if (sgn(dist(l.s, l.e)) == 0)
        return false;
    for (int i = 0; i < v.size(); i++)
        if (!Seg_inter_line(l, v[i]))
            return false;
    return true;
}
int main()
{
    int T, n;
    scanf("%d", &T);
    while (T--)
    {
        v.clear();
        double x1, y1, x2, y2;
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
        {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            v.push_back(Line(Point(x1, y1), Point(x2, y2)));
        }
        bool f = false;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (check(Line(v[i].s, v[j].s)) || check(Line(v[i].s, v[j].e))
                    || check(Line(v[i].e, v[j].s)) || check(Line(v[i].e, v[j].e)))
                {
                    f = true;
                    break;
                }
            }
        }
        if (f)
            printf("Yes!
");
        else
            printf("No!
");
    }
    return 0;
}

POJ 3304 segments 线段和直线相交

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