POJ - 1269 Intersecting Lines 两直线的位置关系
2017-10-29 13:22
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#include <cmath>
#include <cstdio>
const double T = 1e-8;
struct Node{
double x, y;
}p1, p2, p3, p4;
double ChaJi(Node a, Node b, Node c){
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
}
void Solve(){
if(ChaJi(p1, p2, p3) == 0 && ChaJi(p1, p2, p4) == 0) //共线;
puts("LINE");
else if((p1.x-p2.x)*(p3.y-p4.y) == (p1.y-p2.y)*(p3.x-p4.x)) // 平行
puts("NONE");
else{ //相交
double a1 = p1.y-p2.y;
double b1 = p2.x-p1.x;
double c1 = p1.x*p2.y-p2.x*p1.y;
double a2 = p3.y-p4.y;
double b2 = p4.x-p3.x;
double c2 =p3.x*p4.y-p4.x*p3.y;
double x = (c1*b2-c2*b1)/(a2*b1-a1*b2);
double y = (a2*c1-a1*c2)/(a1*b2-a2*b1);
printf("POINT %.2f %.2f\n", x, y);
}
}
int main(){
int n;
while(scanf("%d", &n) != EOF){
printf("INTERSECTING LINES OUTPUT\n");
while(n--){
scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y, &p4.x, &p4.y);
Solve();
}
printf("END OF OUTPUT\n");
}
return 0;
}
#include <cstdio>
const double T = 1e-8;
struct Node{
double x, y;
}p1, p2, p3, p4;
double ChaJi(Node a, Node b, Node c){
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
}
void Solve(){
if(ChaJi(p1, p2, p3) == 0 && ChaJi(p1, p2, p4) == 0) //共线;
puts("LINE");
else if((p1.x-p2.x)*(p3.y-p4.y) == (p1.y-p2.y)*(p3.x-p4.x)) // 平行
puts("NONE");
else{ //相交
double a1 = p1.y-p2.y;
double b1 = p2.x-p1.x;
double c1 = p1.x*p2.y-p2.x*p1.y;
double a2 = p3.y-p4.y;
double b2 = p4.x-p3.x;
double c2 =p3.x*p4.y-p4.x*p3.y;
double x = (c1*b2-c2*b1)/(a2*b1-a1*b2);
double y = (a2*c1-a1*c2)/(a1*b2-a2*b1);
printf("POINT %.2f %.2f\n", x, y);
}
}
int main(){
int n;
while(scanf("%d", &n) != EOF){
printf("INTERSECTING LINES OUTPUT\n");
while(n--){
scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y, &p4.x, &p4.y);
Solve();
}
printf("END OF OUTPUT\n");
}
return 0;
}
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