辣鸡(ljh)

大模拟，按一个点（左下或右上）排序，考虑对他的贡献有四个方向，再减枝即可

第二个样例的图，拿走不谢

#include<cstdio>
#include<iostream>
#include<algorithm>
#define int long long
#define MAXN 100010
using namespace std;
int n;
inline int maxn(int a,int b){
	return a>b?a:b;
}
inline int minn(int a,int b){
	return a<b?a:b;
}
struct rr{
	int x1,x2,y1,y2;
	friend bool operator <(const rr &a,const rr &b){
		if(a.x1==b.x1)return a.y1<b.y1;
		return a.x1<b.x1;
	}
}qw[MAXN];
int ans;
inline int nei(rr a){
	return (a.x2-a.x1)*(a.y2-a.y1)*2;
}
signed main(){
	//freopen("da.in","r",stdin);
	scanf("%lld",&n);
	for(int i=1;i<=n;++i){
		scanf("%lld%lld%lld%lld",&qw[i].x1,&qw[i].y1,&qw[i].x2,&qw[i].y2);
		ans+=nei(qw[i]);
	}
	//cout<<ans<<endl;
	sort(qw+1,qw+n+1);
	int x1,x2,y1,y2,nx1,nx2,ny1,ny2;
	for(int i=1;i<=n;++i){
		x1=qw[i].x1;x2=qw[i].x2;
		y1=qw[i].y1;y2=qw[i].y2;
		//cout<<"QwQ"<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
		for(int j=i+1;j<=n;++j){
			nx1=qw[j].x1;nx2=qw[j].x2;
			ny1=qw[j].y1;ny2=qw[j].y2;
			//cout<<nx1<<" "<<ny1<<" "<<nx2<<" "<<ny2<<endl;
			if(nx1>x2+1)break;
			if(ny1>y2+1)continue;
			if(ny2+1<y1)continue;
			if(ny2+1==y1||ny1==y2+1){
				if(nx1==x2+1){
					++ans;
					continue;
				}
				//cout<<nx1<<" "<<ny1<<" "<<nx2<<" "<<ny2<<endl;
				ans+=2*(minn(x2,nx2)-maxn(x1,nx1)+1);
				if(x1==nx1)--ans;
				if(x2==nx2)--ans;
				//cout<<ans<<endl;
				continue;
			}
			if(nx1==x2+1){
				//cout<<nx1<<" "<<ny1<<" "<<nx2<<" "<<ny2<<endl;
				ans+=2*(minn(y2,ny2)-maxn(y1,ny1)+1);
				if(y1==ny1)--ans;
				if(y2==ny2)--ans;
				//cout<<ans<<endl;
				continue;
			}
		}
	}
	printf("%lld\n",ans);
}