zoukankan      html  css  js  c++  java
  • poj 1269 Intersecting Lines(判断两直线关系,并求交点坐标)

    Intersecting Lines
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 12421   Accepted: 5548

    Description

    We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in a line because they are on top of one another (i.e. they are the same line), 3) intersect in a point. In this problem you will use your algebraic knowledge to create a program that determines how and where two lines intersect.
    Your program will repeatedly read in four points that define two lines in the x-y plane and determine how and where the lines intersect. All numbers required by this problem will be reasonable, say between -1000 and 1000.

    Input

    The first line contains an integer N between 1 and 10 describing how many pairs of lines are represented. The next N lines will each contain eight integers. These integers represent the coordinates of four points on the plane in the order x1y1x2y2x3y3x4y4. Thus each of these input lines represents two lines on the plane: the line through (x1,y1) and (x2,y2) and the line through (x3,y3) and (x4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (x3,y3) and (x4,y4).

    Output

    There should be N+2 lines of output. The first line of output should read INTERSECTING LINES OUTPUT. There will then be one line of output for each pair of planar lines represented by a line of input, describing how the lines intersect: none, line, or point. If the intersection is a point then your program should output the x and y coordinates of the point, correct to two decimal places. The final line of output should read "END OF OUTPUT".

    Sample Input

    5
    0 0 4 4 0 4 4 0
    5 0 7 6 1 0 2 3
    5 0 7 6 3 -6 4 -3
    2 0 2 27 1 5 18 5
    0 3 4 0 1 2 2 5
    

    Sample Output

    INTERSECTING LINES OUTPUT
    POINT 2.00 2.00
    NONE
    LINE
    POINT 2.00 5.00
    POINT 1.07 2.20
    END OF OUTPUT
    

    Source

    题意:给定 1 - 10组直线,判断每组直线的关系,若相交 输出交点坐标,保留两位小数;若平行,输出‘NONE’;若重合,输出‘LINE’;

    输出格式详见标准输出。

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cmath>
     4 #include <cstdlib>
     5 #include <cstring>
     6 #include <math.h>
     7 #include <algorithm>
     8 #include <cctype>
     9 #include <string>
    10 #include <map>
    11 #include <set>
    12 #define ll long long
    13 using namespace std;
    14 const double eps = 1e-8;
    15 int sgn(double x)
    16 {
    17     if(fabs(x) < eps)return 0;
    18     if(x < 0) return -1;
    19     else return 1;
    20 }
    21 struct Point
    22 {
    23     double x,y;
    24     Point(){}
    25     Point(double _x,double _y)
    26     {
    27         x = _x;y = _y;
    28     }
    29     Point operator -(const Point &b)const
    30     {
    31         return Point(x - b.x,y - b.y);
    32     }
    33     double operator ^(const Point &b)const
    34     {
    35         return x*b.y - y*b.x;
    36     }
    37     double operator *(const Point &b)const
    38     {
    39         return x*b.x + y*b.y;
    40     }
    41 };
    42 
    43 struct Line
    44 {
    45     Point s,e;
    46     Line(){}
    47     Line(Point _s,Point _e)
    48     {
    49         s = _s;e = _e;
    50     }
    51     pair<Point,int> operator &(const Line &b)const
    52     {
    53         Point res = s;
    54         if(sgn((s-e)^(b.s-b.e)) == 0)
    55         {
    56             if(sgn((b.s-s)^(b.e-s)) == 0)
    57                 return make_pair(res,0);//两直线重合
    58             else return make_pair(res,1);//两直线平行
    59         }
    60         double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
    61         res.x += (e.x - s.x)*t;
    62         res.y += (e.y - s.y)*t;
    63         return make_pair(res,2);//有交点
    64     }
    65 };
    66 
    67 int main(void)
    68 {
    69     int t;
    70     double x1,x2,x3,x4,y1,y2,y3,y4;
    71     scanf("%d",&t);
    72     printf("INTERSECTING LINES OUTPUT
    ");
    73     while(t--)
    74     {
    75         scanf("%lf %lf %lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
    76         Line l1 = Line(  Point(x1,y1) ,Point(x2,y2)  );
    77         Line l2 = Line(  Point(x3,y3) ,Point(x4,y4)  );
    78         pair<Point,int> ans = l1 & l2;
    79         if(ans.second == 2) printf("POINT %.2f %.2f
    ",ans.first.x,ans.first.y);
    80         else if(ans.second == 0) printf("LINE
    ");
    81         else printf("NONE
    ");
    82     }
    83     printf("END OF OUTPUT
    ");
    84 
    85     return 0;
    86 }
  • 相关阅读:
    GridControl 绑定非绑定列
    AutoMapper
    ABP-TaskEver 路线图
    Angular--页面间切换及传值的四种方法
    知行合一,止于至善
    AutoFac 依赖注入--问题集
    【入门】依赖注入(DI)-开始
    利用JAVA生成二维码
    sqlserver如何关闭死锁进程.
    Hadoop/HBase 配置snappy压缩
  • 原文地址:https://www.cnblogs.com/henserlinda/p/4736099.html
Copyright © 2011-2022 走看看