POJ 1269 Intersecting Lines (两直线位置关系+求交点)
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2022-03-30 09:01:28
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题意:给出两直线位置,求是平行、重合还是有交点,若有交点,输出。
题解:两直线位置关系+求交点
套板子。
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
using namespace std;
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//Compares a double to zero
int sgn(double x) {
if (fabs(x) < eps) return 0;
if (x < 0) return -1;
else return 1;
}
//POINT
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;
}
};
//LINE
struct Line {
Point s, e;
Line() {}
Line(Point _s, Point _e) {
s = _s;
e = _e;
}
//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 n, xa, ya, xb, yb, xc, yc, xd, yd;
int main() {
scanf("%d", &n);
puts("INTERSECTING LINES OUTPUT");
for (int i = 1; i <= n; i++) {
scanf("%d%d%d%d%d%d%d%d", &xa, &ya, &xb, &yb, &xc, &yc, &xd, &yd);
Line line = Line(Point(xa, ya), Point(xb, yb));
Line line2 = Line(Point(xc, yc), Point(xd, yd));
int temp = line.linecrossline(line2);
if (temp == 0) puts("NONE");
else if (temp == 1) puts("LINE");
else {
Point ans = line.crosspoint(line2);
printf("POINT %.2f %.2f\n", ans.x, ans.y);
}
}
puts("END OF OUTPUT");
return 0;
}
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