Segments---poj3304（判断直线与线段的位置关系）

Posted 2020-08-14 西瓜不懂柠檬的酸

题目链接：http://poj.org/problem?id=3304

题意：给你n个线段，求是否有一条直线与所有的线段都相交，有Yes，没有No;

枚举所有的顶点作为直线的两点，然后判断这条直线是否和所有的线段相交即可;注意不能找两个相同的点作为直线上的两点；

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<stdio.h>
#include<queue>
using namespace std;
#define met(a, b) memset(a, b, sizeof(a))
#define mod 1000000007
typedef long long LL;
const int N = 5050;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;

struct point
{
    double x, y;
    point(double x=0, double y=0) : x(x), y(y) {}
    friend point operator -(point p1, point p2)
    {
        return point(p1.x-p2.x, p1.y-p2.y);
    }
    friend int operator ^(point p1, point p2)
    {
        double k = p1.x*p2.y - p1.y*p2.x;

        if(k > eps) return 1;
        if(fabs(k) < eps) return 0;
        return -1;
    }
    friend int operator ==(point p1, point p2)
    {
        return (p1.x == p2.x && p1.y == p2.y);
    }
}p[N];

struct line
{
    point s, e;
    line(point s=0, point e=0) : s(s), e(e) {}
}Line[N];

int Judge(line l, line L[], int n)
{
    for(int i=1; i<=n; i++)
    {
        int k = abs( ((l.s-l.e)^(L[i].s-l.e)) + ((l.s-l.e)^(L[i].e-l.e)) );
        ///判断直线l是否与线段L[i]相交;
        if(k == 2) return 0;///不相交;
    }
    return 1;
}

int main()
{
    int T, n;
    scanf("%d", &T);
    while(T--)
    {
        met(p, 0);
        met(Line, 0);

        int k = 0;
        scanf("%d", &n);
        for(int i=1; i<=n; i++)
        {
            double x1, x2, y1, y2;
            scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
            p[k++] = point(x1, y1);
            p[k++] = point(x2, y2);
            Line[i] = line(p[k-2], p[k-1]);
        }
        int flag = 0;
        for(int i=0; i<k && !flag; i++)
        {
            for(int j=0; j<i && !flag; j++)
            {
                if(p[i] == p[j]) continue;
                flag = Judge(line(p[i], p[j]), Line, n);
            }
        }
        if(flag) puts("Yes!");
        else puts("No!");
    }
    return 0;
}

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