Codeforces Round #635 (Div. 2) D.Xenia and Colorful Gems

题目链接
Xenia is a girl being born a noble. Due to the inflexibility and harshness of her family, Xenia has to find some ways to amuse herself.

Recently Xenia has bought nr red gems, ng green gems and nb blue gems. Each of the gems has a weight.

Now, she is going to pick three gems.

Xenia loves colorful things, so she will pick exactly one gem of each color.

Xenia loves balance, so she will try to pick gems with little difference in weight.

Specifically, supposing the weights of the picked gems are x, y and z, Xenia wants to find the minimum value of ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 (x−y)^2+(y−z)^2+(z−x)^2 (x−y)2+(y−z)2+(z−x)2. As her dear friend, can you help her?

Input

The first line contains a single integer t (1≤t≤100) — the number of test cases. Then t test cases follow.

The first line of each test case contains three integers n r , n g , n b n_r,n_g,n_b nr,ng,nb ( 1 ≤ n r , n g , n b ≤ 1 e 5 ) (1≤n_r,n_g,n_b≤1e5) (1≤nr,ng,nb≤1e5) — the number of red gems, green gems and blue gems respectively.

The second line of each test case contains nr integers r 1 , r 2 , … , r n r ( 1 ≤ r i ≤ 1 e 9 ) r_1,r_2,…,r_{n_{r}} (1≤r_i≤1e9) r1,r2,…,rnr(1≤ri≤1e9) — ri is the weight of the i-th red gem.

The third line of each test case contains ng integers g 1 , g 2 , … , g n g ( 1 ≤ g i ≤ 1 e 9 ) g_1,g_2,…,g_{n_g} (1≤g_i≤1e9) g1,g2,…,gng(1≤gi≤1e9) — gi is the weight of the i-th green gem.

The fourth line of each test case contains nb integers b 1 , b 2 , … , b n b ( 1 ≤ b i ≤ 1 e 9 ) b_1,b_2,…,b_{n_b} (1≤bi≤1e9) b1,b2,…,bnb(1≤bi≤1e9) — bi is the weight of the i-th blue gem.

It is guaranteed that ∑ n r ≤ 1 0 5 , ∑ n g ≤ 1 0 5 , ∑ n b ≤ 1 0 5 ∑n_r≤10^5, ∑n_g≤10^5, ∑n_b≤10^5 ∑nr≤105,∑ng≤105,∑nb≤105 (the sum for all test cases).

Output

For each test case, print a line contains one integer — the minimum value which Xenia wants to find.

Example

input

output

14
1999999996000000002
24
24
14

这题感觉比C好想一点，我一下就想到了记录下标，即对所有的 r , g , b r,g,b r,g,b 预处理，求出与之最近的其他两个位置的下标，打个比方对下标为 i i i 的 r r r，左边最近的 g g g 下标为 p o s 1 pos1 pos1，最近的 b b b 下标为 p o s 2 pos2 pos2，右边最近的 g g g 下标为 p o s 3 pos3 pos3，最近的 b b b 下标为 p o s 4 pos4 pos4，那么 a n s = min ⁡ ( c a l c u l a t e ( a [ p o s 1 ] , a [ i ] , a [ p o s 4 ] ) , c a l c u l a t e ( a [ p o s 2 ] , a [ i ] , a [ p o s 3 ] ) ) ans=\min(calculate(a[pos1],a[i],a[pos4]),calculate(a[pos2],a[i],a[pos3])) ans=min(calculate(a[pos1],a[i],a[pos4]),calculate(a[pos2],a[i],a[pos3]))，
c a l c u l a t e ( x , y , z ) calculate(x,y,z) calculate(x,y,z) 表示计算最小值的函数，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+5;struct node{int k,id;
}p[3*N];ll calculate(int i,int j,int k){return ll(i-j)*ll(i-j)+ll(i-k)*ll(i-k)+ll(j-k)*ll(j-k);
}bool cmp(node a,node b){return a.k<b.k;
}int main(){int t;cin>>t;while(t--){int n1,n2,n3,n;cin>>n1>>n2>>n3;n=n1+n2+n3;int l12[n],l13[n],r12[n],r13[n],l21[n],l23[n],r21[n],r23[n],l31[n],l32[n],r31[n],r32[n];for(int i=0;i<n1;i++) p[i].id=1,cin>>p[i].k;for(int i=n1;i<n1+n2;i++) p[i].id=2,cin>>p[i].k;for(int i=n1+n2;i<n1+n2+n3;i++) p[i].id=3,cin>>p[i].k;sort(p,p+n,cmp);int pos1=-1,pos2=-1,pos3=-1;for(int i=0;i<n;i++){if(p[i].id==1) pos1=i,l12[i]=pos2,l13[i]=pos3;if(p[i].id==2) pos2=i,l21[i]=pos1,l23[i]=pos3;if(p[i].id==3) pos3=i,l31[i]=pos1,l32[i]=pos2;}pos1=-1;pos2=-1,pos3=-1;for(int i=n-1;i>=0;i--){if(p[i].id==1) pos1=i,r12[i]=pos2,r13[i]=pos3;if(p[i].id==2) pos2=i,r21[i]=pos1,r23[i]=pos3;if(p[i].id==3) pos3=i,r31[i]=pos1,r32[i]=pos2;}ll ans=pow(2,63)-1;for(int i=0;i<n;i++){if(p[i].id==1){if(l12[i]!=-1 && r13[i]!=-1) ans=min(ans,calculate(p[l12[i]].k,p[i].k,p[r13[i]].k));if(l13[i]!=-1 && r12[i]!=-1) ans=min(ans,calculate(p[l13[i]].k,p[i].k,p[r12[i]].k));}else if(p[i].id==2){if(l21[i]!=-1 && r23[i]!=-1) ans=min(ans,calculate(p[l21[i]].k,p[i].k,p[r23[i]].k));if(l23[i]!=-1 && r21[i]!=-1) ans=min(ans,calculate(p[l23[i]].k,p[i].k,p[r21[i]].k));}else{if(l31[i]!=-1 && r32[i]!=-1) ans=min(ans,calculate(p[l31[i]].k,p[i].k,p[r32[i]].k));if(l32[i]!=-1 && r31[i]!=-1) ans=min(ans,calculate(p[l32[i]].k,p[i].k,p[r31[i]].k));}}cout<<ans<<endl;}
}