给出平面上两条线段的两个端点,判断这两条线段是否相交(有一个公共点或有部分重合认为相交)。 如果相交,输出"Yes",否则输出"No"。
Input
第1行:一个数T,表示输入的测试数量(1 <= T <= 1000)
第2 - T + 1行:每行8个数,x1,y1,x2,y2,x3,y3,x4,y4。(-10^8 <= xi, yi <= 10^8)
(直线1的两个端点为x1,y1 | x2, y2,直线2的两个端点为x3,y3 | x4, y4)
Output
输出共T行,如果相交输出"Yes",否则输出"No"。
Sample Input
2 1 2 2 1 0 0 2 2 -1 1 1 1 0 0 1 -1
Sample Output
Yes No
#include<stdio.h>
double max(double a,double b)
{
return (a>b)? a:b;
}
double min(double a,double b)
{
return (a<b)? a:b;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
double x1,x2,x3,x4,y1,y2,y3,y4;
scanf("%lf %lf %lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
if(max(x1,x2)<min(x3,x4)||max(x3,x4)<min(x1,x2)||max(y1,y2)<min(y3,y4)||max(y3,y4)<min(y1,y2))
{
printf("NO\n");
}
else
{
double a,b,c,d;
a=(x1-x3)*(y4-y3)-(y1-y3)*(x4-x3);
b=(x2-x3)*(y4-y3)-(x4-x3)*(y2-y3);
c=(x4-x2)*(y1-y2)-(x1-x2)*(y4-y2);
d=(x3-x2)*(y1-y2)-(x1-x2)*(y3-y2);
if(a*b<=0.0&&c*d<=0.0)
printf("Yes\n");
else
printf("NO\n");
}
}
return 0;
}