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. Android--动态添加控件

            [html]      [html]   package com.mrzhu.edittest;      import android.app.Activity;   import ...

  2. HDU 4998 (点的旋转) Rotate

    为了寻找等效旋转操作,我们任选两个点P0和Q0,分别绕这n个点旋转一定的角度后最终得到Pn和Qn 然后已知:P0和Pn共圆,Q0和Qn共圆.所以要找的等效旋转点就是这两个线段的垂直平分线交点O. 等效 ...

  3. HDU 4513 吉哥系列故事——完美队形II

    变形的Manacher算法,在扩展的时候要加入限制条件,满足题目中说的从左到中间身高不减. 其他地方倒是没有什么改动.. //#define LOCAL #include <iostream&g ...

  4. GUI for git|SourceTree|入门基础

    原文链接:http://www.jianshu.com/p/be9f0484af9d 目录 SourceTree简介 SourceTree基本使用 SourceTree&Git部分名词解释 相 ...

  5. 如何使用SQL Server链接服务器访问DB2 Server

    首先,需要安装Microsoft OLE DB Provider for DB2 下载地址:http://download.microsoft.com/download/B/B/2/BB22098A- ...

  6. SQL语句方法语法总结(二)

    1.给表插入数据. (1)INSERT INTO TBL_NAME VALUES (VALUE_1,VALUE_2,...) (2)INSERT INTO TBL_NAME (COL_1,COL_2, ...

  7. HDU 5375 Gray code 格雷码(水题)

    题意:给一个二进制数(包含3种符号:'0'  '1'  '?'  ,问号可随意 ),要求将其转成格雷码,给一个序列a,若转成的格雷码第i位为1,则得分+a[i].求填充问号使得得分最多. 思路:如果了 ...

  8. (转)hadoop基本操作命令

    http://www.cnblogs.com/gpcuster/archive/2010/06/04/1751538.html 在这篇文章中,我们默认认为Hadoop环境已经由运维人员配置好直接可以使 ...

  9. 定时组件quartz系列<二>quartz的原理

    Quartz是一个大名鼎鼎的Java版开源定时调度器,功能强悍,使用方便.   一.核心概念   Quartz的原理不是很复杂,只要搞明白几个概念,然后知道如何去启动和关闭一个调度程序即可.   1. ...

  10. Struts2实现Preparable接口和【struts2】继承ActionSupport类

    Struts2实现Preparable接口 实现preparable接口,实现public void prepare() throws Exception 方法.当你访问某问action指定方法之前, ...