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  • 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";
    }
    
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  • 原文地址:https://www.cnblogs.com/iss-ue/p/12950345.html
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