Codeforces 685C - Optimal Point(分類討論+亂搞)
阿新 • • 發佈:2021-07-11
分類討論+亂搞
\[\begin{cases}
x_i+y_i+z_i-mid\le x_0+y_0+z_0\le x_i+y_i+z_i+mid\\
-x_i+y_i+z_i-mid\le -x_0+y_0+z_0\le -x_i+y_i+z_i+mid\\
x_i-y_i+z_i-mid\le x_0-y_0+z_0\le x_i-y_i+z_i+mid\\
x_i+y_i-z_i-mid\le x_0+y_0-z_0\le x_i+y_i-z_i+mid
\end{cases}
\] ,\(B_{l},B_r,C_l,C_r,D_l,D_r\) 也同理,那麼顯然上述柿子可以等效於:
\[\begin{cases}
A_l\le x_0+y_0+z_0\le A_r\\
B_l\le -x_0+y_0+z_0\le B_r\\
C_l\le x_0-y_0+z_0\le C_r\\
D_l\le x_0+y_0+z_0\le D_r
\end{cases}
\] 。我們還可以注意到 \(x_0+y_0+z_0=a+b+c\),因此柿子又轉化為
\[\begin{cases}
A_l\le a+b+c\le A_r\\
B_l\le a\le B_r\\
C_l\le b\le C_r\\
D_l\le c\le D_r
\end{cases}
\]
分類討論神題。
首先看到最大值最小,一眼二分答案,於是問題轉化為判定性問題,即是否 \(\exists x_0,y_0,z_0\) 滿足 \(\forall i,|x_0-x_i|+|y_0-y_i|+|z_0-z_i|\le mid\)。
柿子中帶個絕對值,不好直接轉化。不過注意到對於任意實數 \(x\) 都有 \(x\le |x|,-x\le |x|\),因此 \(|x-y|\le v\) 的充要條件即是 \(x-y\le v,y-x\le v\),因此上式可以把絕對值拆開得到:
\[\begin{cases} x_0-x_i+y_0-y_i+z_0-z_i\le mid\\ x_0-x_i+y_0-y_i+z_i-z_0\le mid\\ x_0-x_i+y_i-y_0+z_0-z_i\le mid\\ x_0-x_i+y_i-y_0+z_i-z_0\le mid\\ x_i-x_0+y_0-y_i+z_0-z_i\le mid\\ x_i-x_0+y_0-y_i+z_i-z_0\le mid\\ x_i-x_0+y_i-y_0+z_0-z_i\le mid\\ x_i-x_0+y_i-y_0+z_i-z_0\le mid \end{cases} \]將上式整理一下可以得到:
記 \(A_{l}=\max\limits_{i=1}^nx_i+y_i+z_i-mid,A_{r}=\min\limits_{i=1}^nx_i+y_i+z_i-mid\)
注意到這四個不等式中間一項都帶三個未知數,不好處理,不過這三個未知數之間又存在某種聯絡,因此考慮記 \(a=-x_0+y_0+z_0,b=x_0-y_0+z_0,c=x_0+y_0-z_0\),那麼有 \(x_0=\dfrac{b+c}{2},y_0=\dfrac{a+c}{2},z_0=\dfrac{a+b}{2}\)
當然這裡有一個 restriction 是 \(x_0,y_0,z_0\) 必須都是整數,因此必須有 \(a\equiv b\equiv c\pmod{2}\)。故考慮列舉 \(r=a\bmod 2\),記 \(a'=\dfrac{a-r}{2},b'=\dfrac{b-r}{2},c'=\dfrac{c-r}{2}\),那麼上式又變為:
\[\begin{cases} \lceil\dfrac{A_l-3r}{2}\rceil\le a'+b'+c'\le\lfloor\dfrac{A_r-3r}{2}\rfloor\\ \lceil\dfrac{B_l-r}{2}\rceil\le a'\le\lfloor\dfrac{B_r-r}{2}\rfloor\\ \lceil\dfrac{C_l-r}{2}\rceil\le b'\le\lfloor\dfrac{C_r-r}{2}\rfloor\\ \lceil\dfrac{D_l-r}{2}\rceil\le c'\le\lfloor\dfrac{D_r-r}{2}\rfloor \end{cases} \]這個隨便求一求就行了,我相信即便剛學過 OI 的應該也會罷。
時間複雜度 \(n\log A\),其中 \(A=\max\{a_i\}\)
const int MAXN=1e5;
const ll INF=7e18;
int n;ll x[MAXN+5],y[MAXN+5],z[MAXN+5],X,Y,Z;
ll down(ll x){return (x>=0)?(x>>1):(-(-x+1>>1));}//\lfloor x/2 \rfloor
ll up(ll x){return (x>=0)?(x+1>>1):(-(-x>>1));}//\lceil x/2 \rceil
bool check(ll mid){
ll al=-INF,ar=INF,bl=-INF,br=INF;
ll cl=-INF,cr=INF,dl=-INF,dr=INF;
for(int i=1;i<=n;i++){
chkmax(al,x[i]+y[i]+z[i]-mid);chkmin(ar,x[i]+y[i]+z[i]+mid);
chkmax(bl,-x[i]+y[i]+z[i]-mid);chkmin(br,-x[i]+y[i]+z[i]+mid);
chkmax(cl,x[i]-y[i]+z[i]-mid);chkmin(cr,x[i]-y[i]+z[i]+mid);
chkmax(dl,x[i]+y[i]-z[i]-mid);chkmin(dr,x[i]+y[i]-z[i]+mid);
}
for(int r=0;r<2;r++){
ll wl=up(al-3*r),wr=down(ar-3*r);
ll xl=up(bl-r),xr=down(br-r);
ll yl=up(cl-r),yr=down(cr-r);
ll zl=up(dl-r),zr=down(dr-r);
if(wl<=wr&&xl<=xr&&yl<=yr&&zl<=zr&&xl+yl+zl<=wr&&xr+yr+zr>=wl){
ll a=xl,b=yl,c=zl,need=max(0ll,wl-xl-yl-zl);
a+=min(need,xr-xl);need-=min(need,xr-xl);
b+=min(need,yr-yl);need-=min(need,yr-yl);
c+=min(need,zr-zl);need-=min(need,zr-zl);
a=(a<<1|r);b=(b<<1|r);c=(c<<1|r);
X=b+c>>1;Y=c+a>>1;Z=a+b>>1;return 1;
}
} return 0;
}
void solve(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%lld%lld%lld",&x[i],&y[i],&z[i]);
ll l=0,r=INF>>1,p=-1;
while(l<=r){
ll mid=l+r>>1;
if(check(mid)) p=mid,r=mid-1;
else l=mid+1;
} check(p);//printf("%lld\n",p);
printf("%lld %lld %lld\n",X,Y,Z);
}
int main(){
int qu;scanf("%d",&qu);
while(qu--) solve();
return 0;
}