原题链接

题意：

给出四个点，表示两条直线。判断这两条直线的关系：平行，重合，相交(输出交点)

代码：

// Problem: Intersecting Lines
// Contest: Virtual Judge - POJ
// URL: https://vjudge.net/problem/POJ-1269
// Memory Limit: 10 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
//#pragma GCC optimize(1)
//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
//#include<unordered_map>
//#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read(){ll x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')f = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;}

inline void write(ll x){if (x < 0) x = ~x + 1, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');}

#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)

ll ksm(ll a, ll b,ll mod){ll res = 1;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}

const double eps=1e-8;

int sgn(double x){
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}

struct Point{
	double x,y;
	Point(){}
    Point(double _x,double _y){
        x = _x;
        y = _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;
    }
};

struct Line{
	Point s,e;
	bool parallel(Line v){
        return sgn((e-s)^(v.e-v.s)) == 0;
    }
    //`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;
    }
    //`两直线关系`
    //`0 平行`
    //`1 重合`
    //`2 相交`
    int linecrossline(Line v){
        if((*this).parallel(v))
            return v.relation(s)==3;
        return 2;
    }
    //`求两直线的交点`
    //`要保证两直线不平行或重合`
    Point crosspoint(Line v){
        double a1 = (v.e-v.s)^(s-v.s);
        double a2 = (v.e-v.s)^(e-v.s);
        return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
    }
};

int main(){
#ifdef LOCAL
    freopen("in.in","r",stdin);
    freopen("out.out","w",stdout);
#endif
	int _=read,flag=1;
	while(_--){
		double x1,y1,x2,y2,x3,y3,x4,y4;
		cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
		Line a={{x1,y1},{x2,y2}};
		Line b={{x3,y3},{x4,y4}};
		if(flag) puts("INTERSECTING LINES OUTPUT"),flag=0;
		if(a.linecrossline(b)==0) puts("NONE");
		else if(a.linecrossline(b)==1) puts("LINE");
		else{
			cout<<"POINT ";
			Point t=a.crosspoint(b);
			printf("%.2f %.2f\n",t.x,t.y);
		}
		
		
		if(_==0) puts("END OF OUTPUT");
	}
	return 0;
}

标签：ll,return,Point,int,double,POJ1269,Lines,include,Intersecting
来源： https://blog.csdn.net/weixin_45675097/article/details/119898017