Intersection
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 9996   Accepted: 2632

Description

You are to write a program that has to decide whether a given line segment intersects a given rectangle.

An example: 
line: start point: (4,9) 
end point: (11,2) 
rectangle: left-top: (1,5) 
right-bottom: (7,1)

 
Figure 1: Line segment does not intersect rectangle

The line is said to intersect the rectangle if the line and the rectangle have at least one point in common. The rectangle consists of four straight lines and the area in between. Although all input values are integer numbers, valid intersection points do not have to lay on the integer grid.

Input

The input consists of n test cases. The first line of the input file contains the number n. Each following line contains one test case of the format: 
xstart ystart xend yend xleft ytop xright ybottom

where (xstart, ystart) is the start and (xend, yend) the end point of the line and (xleft, ytop) the top left and (xright, ybottom) the bottom right corner of the rectangle. The eight numbers are separated by a blank. The terms top left and bottom right do not imply any ordering of coordinates.

Output

For each test case in the input file, the output file should contain a line consisting either of the letter "T" if the line segment intersects the rectangle or the letter "F" if the line segment does not intersect the rectangle.

Sample Input

1
4 9 11 2 1 5 7 1

Sample Output

F

Source

 
 
 
给了一个线段和矩形。
 
如果线段和矩形的边相交,或者线段在矩形内。输出T
否则输出F
 
/************************************************************
* Author : kuangbin
* Email : kuangbin2009@126.com
* Last modified : 2013-07-15 10:14
* Filename : POJ1410Intersection.cpp
* Description :
* *********************************************************/ #include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h> using namespace std; const double eps = 1e-;
int sgn(double x)
{
if(fabs(x) < eps)return ;
if(x < )return -;
else return ;
}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y)
{
x = _x;y = _y;
}
Point operator -(const Point &b)const
{
return Point(x - b.x,y - b.y);
}
//叉积
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
//点积
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
//绕原点旋转角度B(弧度值),后x,y的变化
void transXY(double B)
{
double tx = x,ty = y;
x = tx*cos(B) - ty*sin(B);
y = tx*sin(B) + ty*cos(B);
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
//两直线相交求交点
//第一个值为0表示直线重合,为1表示平行,为0表示相交,为2是相交
//只有第一个值为2时,交点才有意义
pair<int,Point> operator &(const Line &b)const
{
Point res = s;
if(sgn((s-e)^(b.s-b.e)) == )
{
if(sgn((s-b.e)^(b.s-b.e)) == )
return make_pair(,res);//重合
else return make_pair(,res);//平行
}
double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
res.x += (e.x-s.x)*t;
res.y += (e.y-s.y)*t;
return make_pair(,res);
}
}; //判断线段相交
bool inter(Line l1,Line l2)
{
return
max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e)) <= &&
sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e)) <= ;
} //判断点在线段上
//判断点在线段上
bool OnSeg(Point P,Line L)
{
return
sgn((L.s-P)^(L.e-P)) == &&
sgn((P.x - L.s.x) * (P.x - L.e.x)) <= &&
sgn((P.y - L.s.y) * (P.y - L.e.y)) <= ;
}
//判断点在凸多边形内
//点形成一个凸包,而且按逆时针排序(如果是顺时针把里面的<0改为>0)
//点的编号:0~n-1
//返回值:
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inConvexPoly(Point a,Point p[],int n)
{
for(int i = ;i < n;i++)
{
if(sgn((p[i]-a)^(p[(i+)%n]-a)) < )return -;
else if(OnSeg(a,Line(p[i],p[(i+)%n])))return ;
}
return ;
}
//判断点在任意多边形内
//射线法,poly[]的顶点数要大于等于3,点的编号0~n-1
//返回值
//-1:点在凸多边形外
//0:点在凸多边形边界上
//1:点在凸多边形内
int inPoly(Point p,Point poly[],int n)
{
int cnt;
Line ray,side;
cnt = ;
ray.s = p;
ray.e.y = p.y;
ray.e.x = -100000000000.0;//-INF,注意取值防止越界 for(int i = ;i < n;i++)
{
side.s = poly[i];
side.e = poly[(i+)%n]; if(OnSeg(p,side))return ; //如果平行轴则不考虑
if(sgn(side.s.y - side.e.y) == )
continue; if(OnSeg(side.s,ray))
{
if(sgn(side.s.y - side.e.y) > )cnt++;
}
else if(OnSeg(side.e,ray))
{
if(sgn(side.e.y - side.s.y) > )cnt++;
}
else if(inter(ray,side))
cnt++;
}
if(cnt % == )return ;
else return -;
}
int main()
{
int T;
double x1,y1,x2,y2;
scanf("%d",&T);
while(T--)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
Line line = Line(Point(x1,y1),Point(x2,y2));
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
if(x1 > x2)swap(x1,x2);
if(y1 > y2)swap(y1,y2);
Point p[];
p[] = Point(x1,y1);
p[] = Point(x2,y1);
p[] = Point(x2,y2);
p[] = Point(x1,y2);
if(inter(line,Line(p[],p[])))
{
printf("T\n");
continue;
}
if(inter(line,Line(p[],p[])))
{
printf("T\n");
continue;
}
if(inter(line,Line(p[],p[])))
{
printf("T\n");
continue;
}
if(inter(line,Line(p[],p[])))
{
printf("T\n");
continue;
}
if(inConvexPoly(line.s,p,) >= || inConvexPoly(line.e,p,) >= )
{
printf("T\n");
continue;
}
printf("F\n");
}
return ;
}
 
 

POJ 1410 Intersection(判断线段交和点在矩形内)的更多相关文章

  1. poj 1410 Intersection (判断线段与矩形相交 判线段相交)

    题目链接 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12040   Accepted: 312 ...

  2. [POJ 1410] Intersection(线段与矩形交)

    题目链接:http://poj.org/problem?id=1410 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Sub ...

  3. POJ 1410 Intersection (线段和矩形相交)

    题目: Description You are to write a program that has to decide whether a given line segment intersect ...

  4. POJ 1410 Intersection(线段相交&amp;&amp;推断点在矩形内&amp;&amp;坑爹)

    Intersection 大意:给你一条线段,给你一个矩形,问是否相交. 相交:线段全然在矩形内部算相交:线段与矩形随意一条边不规范相交算相交. 思路:知道详细的相交规则之后题事实上是不难的,可是还有 ...

  5. POJ 1410 Intersection (计算几何)

    题目链接:POJ 1410 Description You are to write a program that has to decide whether a given line segment ...

  6. POJ 3304 Segments 基础线段交判断

    LINK 题意:询问是否存在直线,使得所有线段在其上的投影拥有公共点 思路:如果投影拥有公共区域,那么从投影的公共区域作垂线,显然能够与所有线段相交,那么题目转换为询问是否存在直线与所有线段相交.判断 ...

  7. POJ 1410 Intersection --几何,线段相交

    题意: 给一条线段,和一个矩形,问线段是否与矩形相交或在矩形内. 解法: 判断是否在矩形内,如果不在,判断与四条边是否相交即可.这题让我发现自己的线段相交函数有错误的地方,原来我写的线段相交函数就是单 ...

  8. 简单几何(线段相交) POJ 1410 Intersection

    题目传送门 题意:一个矩形和一条线段,问是否有相交 分析:考虑各种情况.坑点:给出的矩形的两个端点是无序的,还有线段完全在矩形内也算相交 /****************************** ...

  9. poj 1410 Intersection 线段相交

    题目链接 题意 判断线段和矩形是否有交点(矩形的范围是四条边及内部). 思路 判断线段和矩形的四条边有无交点 && 线段是否在矩形内. 注意第二个条件. Code #include & ...

随机推荐

  1. jstl的forEach使用和jstl变量实现自增

    <c:forEach items="${reallyChooseSubjectList}" var="reallyChooseSubject"> & ...

  2. UVa 10115 Automatic Editing

    字符串题目就先告一段落了,又是在看balabala不知道在说些什么的英语. 算法也很简单,用了几个库函数就搞定了.本来还担心题里说的replace-by为空的特殊情况需要特殊处理,后来发现按一般情况处 ...

  3. UploadifyAPI-上传插件属性和方法介绍

    上一篇文章简单的介绍了Uploadify上传插件的使用.但是对于常用的属性和方法并没有说明.授人以鱼不如授人以渔,我决定将常用的属性列举出来,供大伙参考参考.           Uploadify属 ...

  4. 深入理解OpenERP的工作流(Workflow)

    一.工作流定义: <?xml version="1.0"?>  <terp><data>    <record model="w ...

  5. linux下编译软件通用方法(memcached为例)

    1)到软件的官网或其他网站下载软件的源码包 2)解压源码包,并切换到源码目录中 3)使用./configure --help查询配置帮助,里面可能会有安装指南(Installation directo ...

  6. VS启用IIS调试的方法及可能碰到的问题。

    经常有这种情况, 开发机本地正常, 但是一旦发布到服务上后, 就出现各种问题. 这是由于开发机和服务器环境不一样造成的, 所以开发时要尽可能的模拟真实性.  这时候, VS的这个功能就帮大忙了. 如何 ...

  7. Windows Tftpd32 DHCP服务器 使用

    /********************************************************************* * Windows Tftpd32 DHCP服务器 使用 ...

  8. location.orgin

    location.orgin 在chrome浏览器下,属性返回的是: 协议(http:).域名.端口(www.cnblogs.com).例如访问http://www.cnblogs.com/,那返回的 ...

  9. list() and tuple()

    >>> l = list('sdfsdf') >>> l ['s', 'd', 'f', 's', 'd', 'f'] >>> t = tuple ...

  10. Xtrabackup流备份与恢复

    Xtrabackup是MySQL数据库的备份不可多得的工具之一.提供了全备,增备,数据库级别,表级别备份等等.最牛X的还有不落盘的备份,即流备份方式.对于服务器上空间不足,或是搭建主从,直接使用流式备 ...