题解 P7014 [CERC2013]Chain & Co.
首先注意到同一维相等的矩形都是不能扣起来的,所以只用考虑
考虑
这两个条件都很好判断。关键在于
正难则反,扫描线判断有无两个区间相离或者包含即可。
其它情况类似,只需将坐标变换一下。时间复杂度
#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rof(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;
const int Maxn=1e6,inf=1e9+7;
inline int read()
{
int x=0,f=1;
char ch=getchar();
while(ch<'0' || ch>'9')
{
if(ch=='-') f=-1;
ch=getchar();
}
while(ch>='0' && ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
int T,n;
struct Square{int x1,y1,z1,x2,y2,z2;};
struct Node{int l,r,op;};
inline bool operator<(Node a,Node b) {return a.l<b.l;}
vector<Square> X,Y,Z;
vector<Node> v;
inline int CheckYZ()
{
if(Y.empty() || Z.empty()) return 0;
int Zmn=inf,Zmx=-inf,Ymn=inf,Ymx=-inf;
for(auto i:Z) Zmn=min(Zmn,i.z1),Zmx=max(Zmx,i.z2);
for(auto i:Y) Ymn=min(Ymn,i.y1),Ymx=max(Ymx,i.y2);
for(auto i:Y) if(i.z1>=Zmn || i.z2<=Zmx) return 0;
for(auto i:Z) if(i.y1>=Ymn || i.y2<=Ymx) return 0;
int yl=-inf,yr=inf,zl=-inf,zr=inf;
v.clear(),v.shrink_to_fit();
for(auto i:Y)
yl=max(yl,i.x1),yr=min(yr,i.x2),
v.push_back((Node){i.x1,i.x2,1});
for(auto i:Z)
zl=max(zl,i.x1),zr=min(zr,i.x2),
v.push_back((Node){i.x1,i.x2,2});
if(yr<=zl || zr<=yl) return 0;
sort(v.begin(),v.end()),yr=zr=-inf;
for(auto i:v)
{
if(i.op==1) {if(zr>=i.r) return 0; yr=max(yr,i.r);}
if(i.op==2) {if(yr>=i.r) return 0; zr=max(zr,i.r);}
}
return 1;
}
inline int CheckXY()
{
if(X.empty() || Y.empty()) return 0;
swap(X,Z);
for(auto &i:X) swap(i.x1,i.z1),swap(i.x2,i.z2);
for(auto &i:Y) swap(i.x1,i.z1),swap(i.x2,i.z2);
for(auto &i:Z) swap(i.x1,i.z1),swap(i.x2,i.z2);
int chk=CheckYZ();
swap(X,Z);
for(auto &i:X) swap(i.x1,i.z1),swap(i.x2,i.z2);
for(auto &i:Y) swap(i.x1,i.z1),swap(i.x2,i.z2);
for(auto &i:Z) swap(i.x1,i.z1),swap(i.x2,i.z2);
return chk;
}
inline int CheckXZ()
{
if(X.empty() || Z.empty()) return 0;
swap(X,Y);
for(auto &i:X) swap(i.x1,i.y1),swap(i.x2,i.y2);
for(auto &i:Y) swap(i.x1,i.y1),swap(i.x2,i.y2);
for(auto &i:Z) swap(i.x1,i.y1),swap(i.x2,i.y2);
int chk=CheckYZ();
swap(X,Y);
for(auto &i:X) swap(i.x1,i.y1),swap(i.x2,i.y2);
for(auto &i:Y) swap(i.x1,i.y1),swap(i.x2,i.y2);
for(auto &i:Z) swap(i.x1,i.y1),swap(i.x2,i.y2);
return chk;
}
inline void Solve()
{
n=read(),X.clear(),Y.clear(),Z.clear();
X.shrink_to_fit(),Y.shrink_to_fit(),Z.shrink_to_fit();
For(i,1,n)
{
int x1=read(),y1=read(),z1=read(),x2=read(),y2=read(),z2=read();
if(x1>x2) swap(x1,x2); if(y1>y2) swap(y1,y2); if(z1>z2) swap(z1,z2);
Square k=(Square){x1,y1,z1,x2,y2,z2};
if(x1==x2) X.push_back(k);
if(y1==y2) Y.push_back(k);
if(z1==z2) Z.push_back(k);
}
if(n==1) {printf("NO\n"); return;}
if((!X.empty())+(!Y.empty())+(!Z.empty())==1) {printf("NO\n"); return;}
int XY=CheckXY(),XZ=CheckXZ(),YZ=CheckYZ();
if((!X.empty())+(!Y.empty())+(!Z.empty())==2)
{
if(XY || YZ || XZ) printf("YES\n");
else printf("NO\n");
return;
}
if((XY && XZ) || (XY && YZ) || (XZ && YZ)) printf("YES\n");
else printf("NO\n");
}
int main()
{
T=read();
while(T--) Solve();
return 0;
}