Intersecting Lines---poj1269(求两直线的位置关系)

Posted 2020-08-15 西瓜不懂柠檬的酸

题目链接：http://poj.org/problem?id=1269

 题意：给你两条直线上的任意不同的两点，然后求两条直线的位置关系，如果相交于一点输出该点坐标;

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<stdio.h>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define met(a, b) memset(a, b, sizeof(a))
#define mod 1000000007
typedef long long LL;
//////////////////////////////////////////////////////////////
const int INF = 0x3f3f3f3f;
const int N = 21;
const double eps = 1e-8;

int Sign(double x)
{
    if(fabs(x) < eps) return 0;
    if(x<0) return -1;
    return 1;
}

struct point
{
    double x, y;
    point(double x_=0, double y_=0):x(x_),y(y_){}
    point operator -(const point &b)const
    {
        return point(x-b.x, y-b.y);
    }
    double operator ^(const point &b)const
    {
        return (x*b.y - b.x*y);
    }
};

struct line
{
    point p1, p2;
    line(){}
    line(point p1_, point p2_):p1(p1_), p2(p2_){}
};
///找到直线L1与L2的交点p0，返回值不同，含义不同；
int FindLinePoint(line L1, line L2, point &p0)
{
    double k = ((L1.p1-L1.p2) ^ (L2.p1-L2.p2));
    if(Sign(k) == 0)
    {
        double t = ((L1.p1-L1.p2)^(L1.p1-L2.p1));
        if(Sign(t) == 0) return 2;///共线；
        return 1;///平行
    }
    p0 = L1.p1;
    double t = ((L1.p1-L2.p1)^(L2.p1-L2.p2))/((L1.p1-L1.p2)^(L2.p1-L2.p2));
    p0.x += (L1.p2.x - L1.p1.x)*t;
    p0.y += (L1.p2.y - L1.p1.y)*t;
    return 3;///相交于一点p0；
}


int main()
{
    printf("INTERSECTING LINES OUTPUT\\n");
    int T;
    scanf("%d", &T);
    while(T--)
    {
        point p0;
        double x1, y1, x2, y2, x3, y3, x4, y4;
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &x1,&y1, &x2,&y2, &x3,&y3, &x4,&y4);
        line L1 = line(point(x1,y1), point(x2,y2));
        line L2 = line(point(x3,y3), point(x4,y4));
        int ans = FindLinePoint(L1, L2, p0);
        if(ans == 1) printf("NONE\\n");///平行；
        if(ans == 2) printf("LINE\\n");///共线；
        if(ans == 3) printf("POINT %.2f %.2f\\n", p0.x, p0.y);
    }
    printf("END OF OUTPUT\\n");
    return 0;
}

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