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uva 10601 - Cubes(置换)

题目链接:uva 10601 - Cubes

题目大意:有12根等长的小木棍,然后每根木棍,输入每根木棍颜色的编号,你的任务是统计出用它们拼出多少种不同的立方体,旋转之后完全相同的立方体被认定相同。

解题思路:polya,然后对应立方体有24种旋转:

  • 不旋转(still):1种,循环长度为12
  • 以对顶点为轴(rot_point):4组,循环长度为3
  • 以对面中心为轴(rot_plane):3组,分别有90,180,270度旋转,分别对应循环长度3,2,3
  • 以对边为轴(rot_edge):6组,除了两条边循环长度为1,其他为2.
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 12;

int u[maxn+5], rod[maxn+5];
ll C[maxn+5][maxn+5];

void init () {
    memset(C, 0, sizeof(C));
    for (int i = 0; i <= maxn; i++) {
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; j++)
            C[i][j] = C[i-1][j-1] + C[i-1][j];
    }
}

ll solve (ll k) {
    int n = 0;
    ll ret = 1;

    for (int i = 0; i < 6; i++) {
        if (u[i] % k)
            return 0;
        u[i] /= k;
        n += u[i];
    }

    for (int i = 0; i < 6; i++) {
        ret *= C[n][u[i]];
        n -= u[i];
    }
    //printf("%lld %lld!!\n", k, ret);
    return ret;
}

ll still () {
    memcpy(u, rod, sizeof(rod));
    return solve(1);
}

ll rot_point () {
    memcpy(u, rod, sizeof(rod));
    return 4 * 2 * solve(3);
}

ll rot_edge () {
    ll ret = 0;
    for (int i = 0; i < 6; i++) {
        for (int j = 0; j < 6; j++) {
            if (rod[i] && rod[j]) {
                memcpy(u, rod, sizeof(rod));
                u[i]--; u[j]--;
                ret += 6 * solve(2);
            }
        }
    }
    return ret;
}

ll rot_plane () {
    ll ret = 0;
    memcpy(u, rod, sizeof(rod));
    ret += solve(4) * 2 * 3;
    memcpy(u, rod, sizeof(rod));
    ret += solve(2) * 3;
    return ret;
}

inline ll polya () {
    return still() + rot_point() + rot_edge() + rot_plane();
}

int main () {
    init();

    int cas, x;
    scanf("%d", &cas);
    while (cas--) {

        memset(rod, 0, sizeof(rod));
        for (int i = 0; i < maxn; i++) {
            scanf("%d", &x);
            rod[x-1]++;
        }

        printf("%lld\n", polya() / 24);
    }
    return 0;
}