2024“钉耙编程”中国大学生算法设计超级联赛（2）1003

绝对不模拟的简单魔方

要相信题目的提示 (直接模拟的代码长达300行) ，由于魔方的特性，不论如何转动脚上的色块颜色不会变动，只要枚举8个角块看看是否一致即可，枚举角块时需确定访问角块颜色的顺序，例如以3号为顶，后左上访问顺序为123即坐标为$(3,4)->(4,3)-(4,4)$ ，那么可以通过此角块的移动来确定其他角块的访问顺序（方向一致即可起点可以变动），存下所有访问顺序和坐标遍历判断即可

#include<bits/stdc++.h>  
  
#define int long long  
using namespace std;  
const int N = 1e4 + 10;  
  
vector<string> s(10);  
  
void solve() {  
    for (int i = 1; i <= 9; i++) {  
        cin >> s[i];  
        s[i] = ' ' + s[i];  
    }  
    map<string, int> able;  
    vector<vector<pair<int, int> > > check(10, vector<pair<int, int>>(3));  
    able["123"]++;  
    able["231"]++;  
    able["312"]++;  
    check[0] = {{3, 4},  
                {4, 3},  
                {4, 4}};  
    able["152"]++;  
    able["521"]++;  
    able["215"]++;  
    check[1] = {{1, 4},  
                {4, 12},  
                {4, 1}};  
    able["145"]++;  
    able["451"]++;  
    able["514"]++;  
    check[2] = {{1, 6},  
                {4, 9},  
                {4, 10}};  
    able["341"]++;  
    able["413"]++;  
    able["134"]++;  
    check[3] = {{3, 6},  
                {4, 6},  
                {4, 7}};  
    able["632"]++;  
    able["326"]++;  
    able["263"]++;  
    check[4] = {{7, 4},  
                {6, 4},  
                {6, 3}};  
    able["436"]++;  
    able["364"]++;  
    able["643"]++;  
    check[5] = {{6, 7},  
                {6, 6},  
                {7, 6}};  
    able["562"]++;  
    able["625"]++;  
    able["256"]++;  
    check[6] = {{6, 1},  
                {6, 12},  
                {9, 4}};  
    able["654"]++;  
    able["546"]++;  
    able["465"]++;  
    check[7] = {{9, 6},  
                {6, 10},  
                {6, 9}};  
    for (int i = 0; i < 8; i++) {  
        char a = s[check[i][0].first][check[i][0].second];  
        char b = s[check[i][1].first][check[i][1].second];  
        char c = s[check[i][2].first][check[i][2].second];  
        string k = "";  
        k += a;  
        k += b;  
        k += c;  
        if (able[k]) continue;  
        else {  
            set<char> ans;  
            ans.insert(a);  
            ans.insert(b);  
            ans.insert(c);  
            for (auto j: ans) {  
                cout << j << ' ';  
            }  
            cout << '\n';  
            return;  
        }  
    }  
    cout << "No problem" << '\n';  
}  
  
  
signed main() {  
    cin.tie(nullptr);  
    cout.tie(nullptr);  
    ios::sync_with_stdio(false);  
    int _ = 1;  
    cin >> _;  
    while (_--) {  
        solve();  
    }  
    return 0;  
}