codeforces 1336 B 思维+二分

碎点

题意：给三个数组，让你从每个数组中各挑一个数满足两两之差的平方和最小。

思路：若想满足两两之和的平方差最小，则满足确定一个数t，其余两个数各是其数组中大于等于t和小于等于t的数。

　　用lower和--upper即可

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '
'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e5+10;
ll ans;
ll pf(ll x)
{
    return x * x;
}
void solve(vector<int>a,vector<int>b,vector<int>c)
{
    for(int x : a){
        auto y = lower_bound(b.begin(),b.end(),x);
        auto z = upper_bound(c.begin(),c.end(),x);
        if(y == b.end() || z == c.begin()) continue;
        z--;
        ll tmp = pf(x-*y) + pf(x-*z) + pf(*y-*z);
        ans = min(ans,tmp);
    }
}
int main()
{
    //freopen("input.txt", "r", stdin);
    int _;
    scanf("%d",&_);
    while(_--)
    {
         ans = 9e18;
        int nr,ng,nb;
        cin>>nr>>ng>>nb;
        std::vector<int>r(nr),g(ng),b(nb);
        rep(i,0,nr-1){
            cin>>r[i];
        }
        rep(i,0,ng-1){
            cin>>g[i];
        }
        rep(i,0,nb-1){
            cin>>b[i];
        }
        sort(r.begin(), r.end());
        sort(g.begin(), g.end());
        sort(b.begin(), b.end());
        
        solve(r,g,b);
        solve(r,b,g);
        solve(g,r,b);
        solve(g,b,r);
        solve(b,r,g);
        solve(b,g,r);
        cout<<ans<<endl;

    }
}