很好的基礎(chǔ)題，判斷直線相交的情況。要注意精度。判斷平行和重合時(shí)，用整數(shù)運(yùn)算比較精確。剩下的事就是解出交點(diǎn)了。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-21 21:31:37
File Name: pku1269.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{ for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) { for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
typedef struct line
{
double a, b, c;
};
typedef struct line_t
{
int x1, y1, x2, y2;
};
typedef struct vec_t
{
int x, y;
};
typedef struct point
{
double x, y;
};
int cross_mul(const vec_t &a, const vec_t &b)
{
return a.x * b.y - a.y * b.x;
}
vec_t vec(const line_t &l)
{
vec_t ret;
ret.x = l.x1 - l.x2;
ret.y = l.y1 - l.y2;
return ret;
}
line make_line(line_t l)
{
line tl;
int sign = 1;
tl.a = l.y2 - l.y1;
if (tl.a < 0)
{
sign = -1;
tl.a = sign * tl.a;
}
tl.b = sign * (l.x1 - l.x2);
tl.c = sign * (l.y1 * l.x2 - l.x1 * l.y2);
return tl;
}
void line_intersect(line l1, line l2, point &p)
{
double d = l1.a * l2.b - l2.a * l1.b;
p.x = (l2.c * l1.b - l1.c * l2.b) / d;
p.y = (l2.a * l1.c - l1.a * l2.c) / d;
}
int main()
{
int ca;
printf("INTERSECTING LINES OUTPUT\n");
for (scanf("%d", &ca); ca--;)
{
line_t l1, l2, l3, l4;
scanf("%d%d%d%d%d%d%d%d", &l1.x1, &l1.y1, &l1.x2, &l1.y2, &l2.x1, &l2.y1, &l2.x2, &l2.y2);
l3.x1 = l1.x1;
l3.y1 = l1.y1;
l3.x2 = l2.x1;
l3.y2 = l2.y1;
l4.x1 = l1.x1;
l4.y1 = l1.y1;
l4.x2 = l2.x2;
l4.y2 = l2.y2;
if (cross_mul(vec(l1), vec(l3)) == 0 && cross_mul(vec(l1), vec(l4)) == 0)
printf("LINE\n");
else if (cross_mul(vec(l1), vec(l2)) == 0)
printf("NONE\n");
else
{
point inter;
line_intersect(make_line(l1), make_line(l2), inter);
printf("POINT %.2lf %.2lf\n", inter.x, inter.y);
}
}
printf("END OF OUTPUT\n");
return 0;
}