Codeforces 464B Restore Cube(暴力)

题目链接：Codeforces 464B Restore Cube

题目大意：给定8个点坐标，对于每个点来说，可以随意交换x，y，z坐标的数值。问说8个点是否可以组成立方体。

解题思路：直接暴力枚举即可，保证一个点的坐标不变，枚举量为67，将上一层判断。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
typedef long long ll;

const int maxn = 8;

struct point {
    ll x[3];

    void input() {
        for (int i = 0; i < 3; i++)
            scanf("%lld", &x[i]);
    }

    void output() {
        printf("%lld %lld %lld\n", x[0], x[1], x[2]);
    }

    point operator - (const point& u) {
        point ret;
        for (int i = 0; i < 3; i++)
            ret.x[i] = x[i] - u.x[i];
        return ret;
    }
}p[maxn+5];

ll dis (point u) {
    ll ret = 0;
    for (int i = 0; i < 3; i++)
        ret += u.x[i] * u.x[i];
    return ret;
}

bool check (int k) {
    int mv = 0;
    ll d[maxn+5];

    for (int i = 0; i < maxn; i++) {
        if (i == k)
            continue;

        d[mv++] = dis(p[k] - p[i]);
    }

    sort(d, d + 7);
    if (d[0] == 0)
        return false;

    if (d[0] != d[1] || d[0] != d[2])
        return false;

    if (d[3] != d[4] || d[3] != d[5])
        return false;

    if (d[0] * 2 != d[3] || d[0] * 3 != d[6])
        return false;
    return true;
}

bool judge () {

    for (int i = 0; i < maxn; i++)
        if (!check(i))
            return false;

    printf("YES\n");
    for (int i = 0; i < maxn; i++)
        p[i].output();
    return true;
}

bool dfs (int d) {

    if (d == maxn)
        return judge();

    sort(p[d].x, p[d].x + 3);
    do {
        if (dfs(d+1))
            return true;
    } while (next_permutation(p[d].x, p[d].x + 3));
    return false;
}

int main () {
    for (int i = 0; i < maxn; i++)
        p[i].input();

    if (!dfs(1))
        printf("NO\n");
    return 0;
}