cf 1336 Xenia and Colorful Gems

cf 1336 Xenia and Colorful Gems

题意 : 给定三组数，分别选择x， y， z，使得 $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2$ 最小。
方法：假设 $x<y<z$ ，给定 x 和 z，y 最接近于$(x+z)/2$时，对于 x，z，值最小。那么 z 接近于 x 的时候，对于 x， 值最小。枚举 x ，选择最接近于 x 的 z，根据 x 和 z选择 y， 二分。
题目链接

代码

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 1e5+10;
const ll INF = 9e18;

int a[N], b[N], c[N];
int la, lb, lc;

int find(int a[], int len, int x) {
    return lower_bound(a, a + len, x) - a;
}

ll cal(int x, int y, int z) {
    return (ll)(x-y) * (x-y) + (ll)(x - z) * (x - z) + (ll)(y-z)*(y-z);
}

ll gett(int x, int z, int b[], int lb) {
    ll res = INF;
    int idx = find(b, lb, x+z >> 1);
    int y;
    if(idx > 0) {
        y = b[idx-1];
        res = min(res, cal(x, y, z));
    }
    if(idx < lb) {
        y = b[idx];
        res = min(res, cal(x, y, z));
    }
    return res;
}

ll solve(int a[], int la, int b[], int lb, int c[], int lc) {
    ll res = INF;
    for(int i = 0; i < la; i++) {
        int x = a[i], y, z;
        int idx = find(c, lc, x);
        if(idx > 0) {
            z = c[idx-1];
            res = min(res, gett(x, z, b, lb));
        }
        if(idx < lc) {
            z = c[idx];
            res = min(res, gett(x, z, b, lb));
        }
    }
    return res;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T -- ) {
        scanf("%d%d%d", &la, &lb, &lc);
        for(int i = 0; i < la; i++) scanf("%d", &a[i]);
        for(int i = 0; i < lb; i++) scanf("%d", &b[i]);
        for(int i = 0; i < lc; i++) scanf("%d", &c[i]);
        sort(a, a + la), sort(b, b + lb), sort(c, c + lc);

        ll ans = INF;
        ans = min(ans, solve(a, la, b, lb, c, lc));
        ans = min(ans, solve(c, lc, b, lb, a, la));

        ans = min(ans, solve(a, la, c, lc, b, lb));
        ans = min(ans, solve(b, lb, c, lc, a, la));

        ans = min(ans, solve(c, lc, a, la, b, lb));
        ans = min(ans, solve(b, lb, a, la, c, lc));

        printf("%lld\n", ans);
    }
    return 0;
}