abc099d<dfs，枚举排列方案>

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
using LL = long long;
const int N = 510, C = 35;
int d[C][C];
int cnt[3][C];
int cost[3][C];
bool vis[C];
int n, m;
int ans = 2e9;

void dfs(int dep, int sum)
{
    if (dep == 3)
    {
        ans = min(ans, sum);
        return;
    }
    for (int i = 1; i <= m; i ++)
        if (vis[i] == false)
        {
            vis[i] = true;
            dfs(dep + 1, sum + cost[dep][i]);
            vis[i] = false;
        }
}

void solv()
{
    
    cin >> n >> m;
    // 转换代价
    for (int i = 1; i <= m; i ++)
        for (int j = 1; j <= m; j ++)
            cin >> d[i][j];
    // 统计三组、每种颜色个数
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= n; j ++)
        {
            int c; cin >> c;
            cnt[(i + j) % 3][c] ++;
        }
    // 计算三组、转换为每种颜色的花费
    for (int i = 0; i < 3; i ++)
    {
        for (int j = 1; j <= m; j ++)
        {
            int t = 0;
            for (int k = 1; k <= m; k ++)
                t += d[k][j] * cnt[i][k];
            cost[i][j] = t;
        }
    }

    // 枚举A(m, 3)
    dfs(0, 0);

    cout << ans << '\n';
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    // cin >> T;
    while (T--)
        solv();
    return 0;
}