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简单几何(直线位置) POJ 1269 Intersecting Lines

时间:2015-10-24 10:05:22      阅读:186      评论:0      收藏:0      [点我收藏+]

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题目传送门

题意:判断两条直线的位置关系,共线或平行或相交

分析:先判断平行还是共线,最后就是相交。平行用叉积判断向量,共线的话也用叉积判断点,相交求交点

 

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/24 星期六 09:08:55
* File Name     :POJ_1269.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
struct Point    {       //点的定义
    double x, y;
    Point (double x=0, double y=0) : x (x), y (y) {}
};
typedef Point Vector;       //向量的定义
Point read_point(void)   {      //点的读入
    double x, y;
    scanf ("%lf%lf", &x, &y);
    return Point (x, y);
}
double polar_angle(Vector A)  {     //向量极角
    return atan2 (A.y, A.x);
}
double dot(Vector A, Vector B)  {       //向量点积
    return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B)    {       //向量叉积
    return A.x * B.y - A.y * B.x;
}
int dcmp(double x)  {       //三态函数,减少精度问题
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
Vector operator + (Vector A, Vector B)  {       //向量加法
    return Vector (A.x + B.x, A.y + B.y);
}
Vector operator - (Vector A, Vector B)  {       //向量减法
    return Vector (A.x - B.x, A.y - B.y);
}
Vector operator * (Vector A, double p)  {       //向量乘以标量
    return Vector (A.x * p, A.y * p);
}
Vector operator / (Vector A, double p)  {       //向量除以标量
    return Vector (A.x / p, A.y / p);
}
bool operator < (const Point &a, const Point &b)    {       //点的坐标排序
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}
bool operator == (const Point &a, const Point &b)   {       //判断同一个点
    return dcmp (a.x - b.x) == 0 && dcmp (a.y - b.y) == 0;
}
double length(Vector A) {       //向量长度,点积
    return sqrt (dot (A, A));
}
double angle(Vector A, Vector B)    {       //向量转角,逆时针,点积
    return acos (dot (A, B) / length (A) / length (B));
}
double area_triangle(Point a, Point b, Point c) {       //三角形面积,叉积
    return fabs (cross (b - a, c - a)) / 2.0;
}
Vector rotate(Vector A, double rad) {       //向量旋转,逆时针
    return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
Vector nomal(Vector A)  {       //向量的单位法向量
    double len = length (A);
    return Vector (-A.y / len, A.x / len);
}
Point point_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点,参数方程
    Vector U = p - q;
    double t = cross (W, U) / cross (V, W);
    return p + V * t;
}
double dis_to_line(Point p, Point a, Point b)   {       //点到直线的距离,两点式
    Vector V1 = b - a, V2 = p - a;
    return fabs (cross (V1, V2)) / length (V1);
}
double dis_to_seg(Point p, Point a, Point b)    {       //点到线段的距离,两点式
  
    if (a == b) return length (p - a);
    Vector V1 = b - a, V2 = p - a, V3 = p - b;
    if (dcmp (dot (V1, V2)) < 0)    return length (V2);
    else if (dcmp (dot (V1, V3)) > 0)   return length (V3);
    else    return fabs (cross (V1, V2)) / length (V1);
}
Point point_proj(Point p, Point a, Point b)   {     //点在直线上的投影,两点式
    Vector V = b - a;
    return a + V * (dot (V, p - a) / dot (V, V));
}
bool inter(Point a1, Point a2, Point b1, Point b2)  {       //判断线段相交,两点式
    double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1),
           c3 = cross (b2 - b1, a1 - b1), c4 = cross (b2 - b1, a2 - b1);
    return dcmp (c1) * dcmp (c2) < 0 && dcmp (c3) * dcmp (c4) < 0;
}
bool on_seg(Point p, Point a1, Point a2)    {       //判断点在线段上,两点式
    return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0;
}
double area_poly(Point *p, int n)   {       //多边形面积
    double ret = 0;
    for (int i=1; i<n-1; ++i)   {
        ret += fabs (cross (p[i] - p[0], p[i+1] - p[0]));
    }
    return ret / 2;
}

int main(void)    {
    int T;  scanf ("%d", &T);
    Point a1, a2, b1, b2;
    puts ("INTERSECTING LINES OUTPUT");
    while (T--) {
        a1 = read_point ();
        a2 = read_point ();
        b1 = read_point ();
        b2 = read_point ();
        Vector A = a2 - a1, B = b2 - b1;
        if (cross (A, B) == 0)   {
            if (cross (b1 - a1, b2 - a1) == 0 && cross (b1 - a2, b2 - a2) == 0)  puts ("LINE");
            else    puts ("NONE");
        }
        else    {
            Point ans = point_inter (a1, a2 - a1, b1, b2 - b1);
            printf ("POINT %.2f %.2f\n", ans.x, ans.y);
        }
    }
    puts ("END OF OUTPUT");

    return 0;
}

  

简单几何(直线位置) POJ 1269 Intersecting Lines

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原文地址:http://www.cnblogs.com/Running-Time/p/4906238.html

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