题意

给你3个数组\(a, b\)和\(c\)，最小化\((x-y)^2+(y-z)^2+(z-x)^2\)，其中\(x \in a, y \in b, z \in c\)。

解题思路

这题其实第一眼就秒了，但是赛中突然不相信自己的直觉，然后就想复杂了。

就很气，本来30min前能过得题被我硬生生拖到了1h+。

解法简单的说，就是固定\(x\)，然后分别在\(b\)和\(c\)中找出\(x\)的前驱和后继，然后两两组合，所有情况的最小值就是答案。

AC代码

#include <bits/stdc++.h>
using namespace std;
 
// #include <ext/rope>
// using namespace __gnu_cxx;
 
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
 
// typedef ll key_type;
// typedef null_mapped_type value_type;
// typedef tree<key_type, value_type, less<key_type>, rb_tree_tag, tree_order_statistics_node_update> rbtree;
 
// typedef __gnu_pbds::priority_queue<pi,greater<pi>,pairing_heap_tag > heap;
 
// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// int rnd(int l,int r){return l+rng()%(r-l+1);}
 
typedef long long ll;
typedef double db;
typedef pair<int,int> PI;
typedef vector<int> VI;
 
#define rep(i,_,__) for (int i=_; i<=__; ++i)
#define per(i,_,__) for (int i=_; i>= __; --i)
 
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define x1 _x
#define x2 __x
#define y1 _y
#define y2 __y
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define endl '\n'
   
const double pi = acos(-1.0);
   
namespace IO{
    bool REOF = 1; //为0表示文件结尾
    inline char nc() {
        static char buf[1 << 20], *p1 = buf, *p2 = buf;
        return p1 == p2 && REOF && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? (REOF = 0, EOF) : *p1++;
    }
   
    template<class T>
    inline bool read(T &x) {
        char c = nc();bool f = 0; x = 0;
        while (c<'0' || c>'9')c == '-' && (f = 1), c = nc();
        while (c >= '0'&&c <= '9')x = (x << 3) + (x << 1) + (c ^ 48), c = nc();
        if(f)x=-x;
        return REOF;
    }
 
    template<class T>
    inline bool write(T x){
        if(x > 9) write(x / 10);
        putchar('0'+x%10);
        return REOF;
    }
   
    template<typename T, typename... T2>
    inline bool read(T &x, T2 &... rest) {
        read(x);
        return read(rest...);
    }
   
   
    inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')); }
    // inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')) || c==' '; }
   
    inline bool read_str(char *a) {
        while ((*a = nc()) && need(*a) && REOF)++a; *a = '\0';
        return REOF;
    }
   
    inline bool read_db(double &x){
        bool f = 0; char ch = nc(); x = 0;
        while(ch<'0'||ch>'9')  {f|=(ch=='-');ch=nc();}
        while(ch>='0'&&ch<='9'){x=x*10.0+(ch^48);ch=nc();}
        if(ch == '.') {
            double tmp = 1; ch = nc();
            while(ch>='0'&&ch<='9'){tmp=tmp/10.0;x=x+tmp*(ch^48);ch=nc();}
        }
        if(f)x=-x;
        return REOF;
    }
   
    template<class TH>
    inline void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<'='<<h<<endl; }
   
    template<class TH, class... TA>
    inline void _dbg(const char *sdbg, TH h, TA... a) {
        while(*sdbg!=',')cerr<<*sdbg++;
        cerr<<'='<<h<<','<<' '; _dbg(sdbg+1, a...);
    }
      
    template<class T>
    ostream &operator<<(ostream& os, vector<T> V) {
        os << "[ "; for (auto vv : V) os << vv << ","; return os << " ]";
    }
   
    template<class T>
    ostream &operator<<(ostream& os, set<T> V) {
        os << "[ "; for (auto vv : V) os << vv << ","; return os << " ]";
    }
 
    template<class T>
    ostream &operator<<(ostream& os, map<T,T> V) {
        os << "[ "; for (auto vv : V) os << vv << ","; return os << " ]";
    }
  
    template<class L, class R>
    ostream &operator<<(ostream &os, pair<L,R> P) {
        return os << "(" << P.x << "," << P.y << ")";
    }
 
    #ifdef BACKLIGHT
    #define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
    #else
    #define debug(...)
    #endif
}
 
 
 
 
using namespace IO;
const int N = 2e5 + 5;
const int M = 2e5 + 5;
const int MAXV = 1e6 + 5;
const int MOD = 1e9+7;              // 998244353 1e9+7
const int INF = 0x3f3f3f3f;             // 1e9+7 0x3f3f3f3f
const ll LLINF = 0x3f3f3f3f3f3f3f3f;    // 1e18+9 0x3f3f3f3f3f3f3f3f
const double eps = 1e-8;
 
// int dx[4] = { 0, 1, 0, -1 };
// int dx[8] = { 1, 0, -1, 1, -1, 1, 0, -1 };
// int dy[4] = { 1, 0, -1, 0 };
// int dy[8] = { 1, 1, 1, 0, 0, -1, -1, -1 };
 
 
 
 
// ll qp(ll a, ll b) {
//     ll res = 1;
//     a %= mod;
//     assert(b >= 0);
//     while(b){
//         if(b&1)
//             res = res * a % mod;
//         a = a * a % mod;
//         b >>= 1;
//     }
//     return res;
// }
// ll inv(ll x) {return qp(x, mod - 2);}
// ll factor[N], finv[N];
// void init() {
//  factor[0]=1;
//  for(int i=1; i<N; i++) factor[i] = factor[i-1] * i % mod;
//  finv[N-1] = qp(factor[N-1], mod - 2);
//  for(int i=N-2; i>=0; i--) finv[i] = finv[i+1] * (i+1) % mod;
// }
// ll c(ll n, ll m) {
//     return factor[n] * finv[m] % mod * finv[n-m] % mod;
// }
 
 
 
 
// #define ls (x<<1)
// #define rs (x<<1|1)
// #define mid ((l+r)>>1)
// #define lson ls,l,mid
// #define rson rs,mid+1,r
 
 
#define fore(_, __) for(int _ = head[__]; _; _=e[_].nxt)
int head[N], tot = 1;
struct Edge {
    int v, nxt;
    Edge(){}
    Edge(int _v, int _nxt):v(_v), nxt(_nxt) {}
}e[N << 1];
void addedge(int u, int v) {
    e[tot] = Edge(v, head[u]); head[u] = tot++;
    e[tot] = Edge(u, head[v]); head[v] = tot++;
}
 
 
   
/**
 * **********     Backlight     **********
 * 仔细读题
 * 注意边界条件
 * 记得注释输入流重定向
 * 没有思路就试试逆向思维
 * 我不打了，能不能把我的分还给我
 */

inline int getpre(vector<int>& v, int val) {
    auto it = upper_bound(all(v), val);
    if(it==v.begin()) return -1;
    return *(--it);
}

inline int getsuf(vector<int>& v, int val) {
    auto it = lower_bound(all(v), val);
    if(it==v.end()) return -1;
    return *it;
}

inline ll calc(ll x, ll y, ll z) {
    ll res = (x-y) * (x-y) + (y-z) * (y-z) + (z-x) * (z-x);
    // debug(x, y, z, res);
    return res;
}

ll work(int A, int B, int C, vector<int>& a, vector<int>& b, vector<int>& c) {
    ll res = LLINF;
    rep(i, 0, A-1) {
        int preb = getpre(b, a[i]);
        int prec = getpre(c, a[i]);
        int sufb = getsuf(b, a[i]);
        int sufc = getsuf(c, a[i]);

        if(preb != -1 && prec != -1) {
            ll tmp = calc(a[i], preb, prec);
            res = min(res, tmp);
        }

        if(preb != -1 && sufc != -1) {
            ll tmp = calc(a[i], preb, sufc);
            res = min(res, tmp);
        }

        if(sufb != -1 && prec != -1) {
            ll tmp = calc(a[i], sufb, prec);
            res = min(res, tmp);
        }

        if(sufb != -1 && sufc != -1) {
            ll tmp = calc(a[i], sufb, sufc);
            res = min(res, tmp);
        }
    }
    // debug(res);
    return res;
}

int A, B, C;
vector<int> a, b, c;
void solve(int Case) {
    read(A, B, C);
    a.resize(A);
    b.resize(B);
    c.resize(C);
    rep(i, 0, A-1) read(a[i]);
    rep(i, 0, B-1) read(b[i]);
    rep(i, 0, C-1) read(c[i]);

    sort(all(a));
    sort(all(b));
    sort(all(c));

    ll ans = work(A, B, C, a, b, c);
    ans = min(ans, work(B, C, A, b, c, a));
    ans = min(ans, work(C, A, B, c, a, b));
    printf("%lld\n", ans);
}
 
 
int main()
{
#ifdef BACKLIGHT
    freopen("in.txt", "r", stdin);
#endif
    // ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int _T; read(_T); for (int _ = 1; _ <= _T; _++) solve(_);
    // int _T=1; while(read(n)) solve(_T), _T++;
    // solve(1);
    return 0;
}