POJ 1269Intersecting Lines [计算几何]

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题目

Intersecting Lines

Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 24321Accepted: 10027

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

解释与代码

kuangbin模板很容易就过了

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <iostream>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <vector>
#include <limits.h>
#include <assert.h>
#include <functional>
#include <numeric>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define pb          push_back
#define ppb         pop_back
#define lbnd        lower_bound
#define ubnd        upper_bound
#define endl        '\\n'
#define trav(a, x)  for(auto& a : x)
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define DEBUG       cerr<<"/n>>>I'm Here<<</n"<<endl;
#define display(x)  trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x)  cerr << #x << " is " << x << endl;
#define ini(a)      memset(a,0,sizeof(a))
#define case        ll T;read(T);for(ll Q=1;Q<=T;Q++)
#define lowbit(x)   x&(-x)
#define pr          printf
#define sc          scanf
//#define _           0
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define DBG(x) \\
    (void)(cout << "L" << __LINE__ \\
    << ": " << #x << " = " << (x) << '\\n')
#define TIE \\
    cin.tie(0);cout.tie(0);\\
    ios::sync_with_stdio(false);
//#define long long int

//using namespace __gnu_pbds;

template <typename T>
void read(T &x) {
    x = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        x = x * 10 + (ch ^ 48);
        ch = getchar();
    }
    x *= f;
    return;
}

inline void write(long long x) {
    if(x<0) putchar('-'), x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
    putchar('\\n');
}

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double pi    = acos(-1.0);
const double eps   = 1e-8;
const int    INF   = 0x3f3f3f3f;
const ll     LLINF = 0x3f3f3f3f3f3f3f3f;
const int    maxn  = 200009;
const ll     N     = 5;

const double inf=1e20;
const int maxp=1010;

//判断正负
int sgn(double x) {
	if (fabs(x)<eps) return 0;
	if (x<0) return -1;
	else return 1;
}
//平方
inline double sqr(double x) {
	return x*x;
}

struct Point {
	double x,y;
	Point() {}
	Point(double _x, double _y) {
		x=_x;
		y=_y;
	}
	void input() {
		scanf("%lf %lf",&x,&y);
	}
	void output() {
		printf("%.2f %.2f\\n",x,y);
	}
	bool operator == (Point b)const {
		return sgn(x-b.x)==0 && sgn(y-b.y)==0;
	}
	bool operator < (Point b)const {
		return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
	}
	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;
	}
	//返回长度
	double len() {
		return hypot(x,y);
	}
	//返回长度平方
	double len2() {
		return x*x+y*y;
	}
	//返回两点间距
	double distance(Point p) {
		return hypot(x-p.x,y-p.y);
	}
	Point operator +(const Point &b)const {
		return Point(x+b.x,y+b.y);
	}
	Point operator *(const double &k)const {
		return Point(x*k,y*k);
	}
	Point operator /(const double &k)const {
		return Point(x/k,y/k);
	}
	//pa和pb的夹角
	double rad(Point a,Point b) {
		Point p=*this;
		return fabs(atan2(fabs((a-p)^(b-p)),(a-p)*(b-p)));
	}
	//化为长度为r的向量
	Point trunc(double r) {
		double l=len();
		if (!sgn(l)) return *this;
		r/=l;
		return Point(x*r,y*r);
	}
	//逆时针旋转 90 度
	Point rotleft() {
		return Point(-y,x);
	}
	//顺时针旋转 90 度
	Point rotright() {
		return Point(y,-x);
	}
	//绕着 p 点逆时针旋转 angle
	Point rotate(Point p,double angle) {
		Point v=(*this)-p;
		double c=cos(angle),s=sin(angle);
		return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
	}
};


struct Line {
	Point s,e;
	Line() {}
	Line(Point _s,Point _e) {
		s=_s;
		e=_e;
	}
	bool operator ==(Line v) {
		return (s==v.s) && (e==v.e);
	}
	//根据一个点和倾斜角 angle 确定直线,0<=angle<π
	Line(Point p, double angle) {
		s=p;
		if (sgn(angle-pi/2)==0) {
			e=(s+Point(0,1));
		} 
		else {
			e=(s+Point(1,tan(angle)));
		}
	}
	//ax+by+c=0
	Line(double a,double b,double c) {
		if(sgn(a)==0) {
			s=Point(0,-c/b);
			e=Point(1,-c/b);
		} 
		else if(sgn(b)==0) {
			s=Point(-c/a,0);
			e=Point(-c/a,1);
		} 
		else {
			s=Point(0,-c/b);
			e=Point(1,(-c-a)/b);
		}
	}
	void input() {
		s.input();
		e.input();
	}
	void adjust() {
		if(e<s) swap(s,e);	
	}
	//求线段长度
	double length() {
		return s.distance(e);
	}
	//返回直线倾斜角 0<=angle<π 
	double angle() {
		double k=atan2(e.y-s.y,e.x-s.x);
		if(sgn(k)<0) k+=pi;
		if(sgn(k-pi)==0) k-= pi;
		return k;
	}
	//点和直线关系
	// 1 在左侧
	// 2 在右侧
	// 3 在直线上 
	int relation(Point p) {
		int c=sgn((p-s)^(e-s));
		if(c<0) return 1; 
		else if(c>0) return 2; 
		else return 3;	 
	}
	//点在线段上的判断
	bool pointonseg(Point p) {
		return sgn((p-s)^(e-s))==0 && sgn((p-s)*(p-e))<=0;
	}
	//两向量平行 (对应直线平行或重合)
	bool parallel(Line v) {
		return sgn((e-s)^(v.e-v.s))==0;
	}
	//两线段相交判断
	//2 规范相交
	//1 非规范相交
	//0 不相交
	int segcrossseg(Line v) {
		int d1=sgn((e-s)^(v.s-s));
		int d2=sgn((e-s)^(v.e-s));
		int d3=sgn((v.e-v.s)^(s-v.s));
		int d4=sgn((v.e-v.s)^(e-v.s));
		if((d1^d2)==-2&&(d3^d4)==-2)return 2;
		return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
		       (d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
		       (d3==0 && sgn((s-v.s)*(s-v.e))●POJ 1269 Intersecting Lines


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