传送门

题意：

给你一个白色的矩形和俩个黑色的矩形，问白色被黑色覆盖后还能不能看到。

思路：

经典被简单题卡。
一开始写了个自我感觉很对的做法，结果$wa41$，检查不出来只能换思路了。
考虑一个容斥，我们算出来第一个黑色矩形覆盖的面积，再算第二个黑色矩形覆盖的面积，让后减去两个覆盖的矩形的面积交。此时就是黑色矩形覆盖白色的总面积，只需要算出白色的面积让后看看是否等于即可。

// Problem: C. White Sheet
// Contest: Codeforces - Codeforces Round #587 (Div. 3)
// URL: https://codeforces.com/contest/1216/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=4000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int x[N],y[N];pair<PII,PII> solve(PII a,PII b,PII c,PII d) {int lx=max(a.X,c.X),rx=min(b.X,d.X);int ly=max(a.Y,c.Y),ry=min(b.Y,d.Y);if(lx>=rx||ly>=ry) return {{0,0},{0,0}};return {{lx,ly},{rx,ry}};
}LL getarea(pair<PII,PII> now) {return 1ll*(now.X.X-now.Y.X)*(now.X.Y-now.Y.Y);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int x1,y1,x2,y2; cin>>x1>>y1>>x2>>y2;int x3,y3,x4,y4; cin>>x3>>y3>>x4>>y4;int x5,y5,x6,y6; cin>>x5>>y5>>x6>>y6;pair<PII,PII> now1=solve({x1,y1},{x2,y2},{x3,y3},{x4,y4});pair<PII,PII> now2=solve({x1,y1},{x2,y2},{x5,y5},{x6,y6});pair<PII,PII> now3=solve(now1.X,now1.Y,now2.X,now2.Y);LL ans1=getarea(now1),ans2=getarea(now2),ans3=getarea(now3);if(getarea({{x1,y1},{x2,y2}})>ans1+ans2-ans3) puts("YES");else puts("NO");return 0;
}
/**/