1216C

(算出a和b的交集q,a和c的交集w)

(再算出q和w的交集e)

(如果q+w-e是a的面积,那么a就被完全覆盖了)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct p{
	ll x1,y1,x2,y2;
}a,b,c;
ll ji(p a,p b)
{
	ll x=min(a.x2,b.x2)-max(a.x1,b.x1);
	ll y=min(a.y2,b.y2)-max(a.y1,b.y1);
	if(x<0||y<0)	return 0;
	else	return x*y;
}
p make_p(p a,p b)
{
	p temp;
	temp.x1=max(a.x1,b.x1);
	temp.x2=min(a.x2,b.x2);
	temp.y1=max(a.y1,b.y1);
	temp.y2=min(a.y2,b.y2);
	return temp;
}
int main()
{
	cin >> a.x1 >> a.y1 >> a.x2 >> a.y2;
    cin >> b.x1 >> b.y1 >> b.x2 >> b.y2;
    cin >> c.x1 >> c.y1 >> c.x2 >> c.y2;
	ll s=ji(a,c)+ji(a,b)-ji(make_p(a,b),make_p(a,c));
	if( s!=(a.x2-a.x1)*(a.y2-a.y1) )	cout<<"YES";
	else	cout<<"NO";
}