NYOJ 1132 promise me a medal （判断两线段是否相交）
promise me a medal
时间限制：
1000 ms | 内存限制：
65535 KB
难度：
2
- 描述
-
you all know that creat2012 will go to a regional. and i want you all pray for us. thx.
for this problem, you need judge if two segments is intersection(相交的). if intersect(相交), print yes and the point. else print no.
easy ? ac it.
- 输入
-
T <= 100 cases
8 real numbers (double)
promise two segments is no coincidence. - 输出
-
no
or yes and two real number (one decimal) - 样例输入
-
2 0 0 2 2 1 0 3 2 0 0 2 2 0 2 2 0
- 样例输出
-
no yes 1.0 1.0
-
坑点在于首尾相连这一情况
思路：先判断线段是否相交，然后在求交点
ac代码：
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define MAXN 201000
#define MAX(a,b) a>b?a:b
#define fab(a) ((a)>0?(a):-(a))
using namespace std;
struct s
{
double x;
double y;
}fb,fe,sb,se,r;
double a,b,c,a1,b1,c1,a2,b2,c2;
void fun(s p,s q)
{
a=q.y-p.y;
b=p.x-q.x;
c=q.x*p.y-p.x*q.y;
}
int find(s p1,s p2,s q1,s q2)
{
double m,n;
fun(p2,q2);
a2=a,b2=b,c2=-c;
m=a*p1.x+b*p1.y+c;
n=a*q1.x+b*q1.y+c;
if(m*n>0)
return 0;
if(fab(m)<1e-8&&fab(n)<1e-8)
{
if(fab(b)<1e-8)
{
if(((p1.y<min(p2.y,q2.y)&&q1.y<min(p2.y,q2.y))||(p1.y>max(p2.y,q2.y)&&q1.y>max(p2.y,q2.y))))
return 0;
else
{
if(p1.y==p2.y||p1.y==q2.y)
r.x=p1.x,r.y=p1.y;
else
r.x=q1.x,r.y=q1.y;
return 1;
}
}
if(((p1.x<min(p2.x,q2.x)&&q1.x<min(p2.x,q2.x))||(p1.x>max(p2.x,q2.x)&&q1.x>max(p2.x,q2.x))))
return 0;
else
{
if(p1.x==p2.x||p1.x==q2.x)
r.x=p1.x,r.y=p1.y;
else
r.x=q1.x,r.y=q1.y;
return 1;
}
}
fun(p1,q1);
a1=a,b1=b,c1=-c;
m=a*p2.x+b*p2.y+c;
n=a*q2.x+b*q2.y+c;
if(m*n>0)
return 0;
r.x=(c1*b2-c2*b1)/(a1*b2-a2*b1);
r.y=(a1*c2-a2*c1)/(a1*b2-a2*b1);
return 1;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&fb.x,&fb.y,&fe.x,&fe.y,&sb.x,&sb.y,&se.x,&se.y);
if(find(fb,sb,fe,se))
printf("yes %.1lf %.1lf\n",r.x,r.y);
else
printf("no\n");
}
return 0;
}