【題目敘述】https://zerojudge.tw/ShowProblem?problemid=c113
【解題想法】
- line-1: (x1, y1) – (x2, y2)，斜率 a1, 截距 b1。
- line-2: (x3, y3) – (x4, y4)，斜率 a2, 截距 b2。
- 加上自己定義的 line-3: (x1, y1) – (x4, y4)，斜率 a3, 截距 b3。
- 如果 a1 = a2 且 a1 = a3，line-1與line-2 重合。
- 如果 a1 = a2 但 a1 != a3，line-1與line-2 平行。
- 其餘情況則計算 line-1與line-2 的交點。需另考慮有一條水平線及一條垂直線的情況。
#include <iostream>
#include <iomanip>
using namespace std;
#define INF 9.9e9
struct Line{
double x1, y1, x2, y2;
double a, b; // y = ax + b
};
void calSlope(Line &l){
if (l.x1 == l.x2){
l.a = INF;
l.b = INF;
} else {
l.a = (l.y2 - l.y1) / (l.x2 - l.x1);
l.b = l.y1 - l.a * l.x1;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int T;
Line l1, l2, l3;
cin >> T;
cout << "INTERSECTING LINES OUTPUT\n";
while (T--){
cin >> l1.x1 >> l1.y1 >> l1.x2 >> l1.y2;
cin >> l2.x1 >> l2.y1 >> l2.x2 >> l2.y2;
l3.x1 = l1.x1; l3.y1 = l1.y1; l3.x2 = l2.x2; l3.y2 = l2.y2;
calSlope(l1);
calSlope(l2);
calSlope(l3);
if (l1.a == l2.a){
if (l1.b == l3.b) cout << "LINE\n";
else cout << "NONE\n";
} else {
double x, y;
if ((l1.x1 == l1.x2) && (l2.y1 == l2.y2)){
x = l1.x1;
y = l2.y2;
} else if ((l2.x1 == l2.x2) && (l1.y1 == l1.y2)){
x = l2.x1;
y = l1.y2;
} else {
x = (l2.b - l1.b) / (l1.a - l2.a);
y = l1.a * x + l1.b;
}
cout << "POINT ";
cout << fixed << setprecision(2)
<< x << " " << y << "\n";
}
}
cout << "END OF OUTPUT\n";
return 0;
}