枚举

【问题描述】

小红给定如下一个 6 × 6 矩阵：

11 8 3 27 24 1

2 21 16 35 17 4

9 29 20 30 5 10

36 33 13 6 23 7

31 14 15 28 12 25

34 19 18 37 22 39

小苯需要从中恰好选择 3 个两两不重叠的 1 × 2 或 2 × 1 子矩形并染红，要求每个

被染红区域内两个数的和都是奇数。

如果两个方案选出的 3 个区域每个都完全相同，则认为它们是同一种方案，选择顺

序不作区分。

#include <iostream>
#include <vector>
#include <cstdint>
using namespace std;
struct Domino {
    uint8_t a, b; 
    Domino(uint8_t a_, uint8_t b_) : a(a_), b(b_) {}
};
bool overlap(const Domino& x, const Domino& y) {
    return (x.a == y.a) || (x.a == y.b) || (x.b == y.a) || (x.b == y.b);
}
int main() {
    int data[6][6] = {
        {11, 8, 3, 27, 24, 1},
        {2, 21, 16, 35, 17, 4},
        {9, 29, 20, 30, 5, 10},
        {36, 33, 13, 6, 23, 7},
        {31, 14, 15, 28, 12, 25},
        {34, 19, 18, 37, 22, 39}
    };
    vector<Domino> dominoes;
    for (int i = 0; i < 6; ++i) {
        for (int j = 0; j < 6; ++j) {
            if (j + 1 < 6 && ((data[i][j] + data[i][j+1]) & 1)) {
                uint8_t a = i * 6 + j;
                uint8_t b = i * 6 + j + 1;
                dominoes.emplace_back(a, b);
            }
            if (i + 1 < 6 && ((data[i][j] + data[i+1][j]) & 1)) {
                uint8_t a = i * 6 + j;
                uint8_t b = (i + 1) * 6 + j;
                dominoes.emplace_back(a, b);
            }
        }
    }
    unsigned int n =dominoes.size();
    int count = 0;
    for(int i=0;i<n-2;i++){
        for(int j=i+1;j<n-1;j++){
            if (overlap(dominoes[i], dominoes[j])) continue;
            for(int k=j+1;k<n;k++){
                if (overlap(dominoes[i], dominoes[k]) || overlap(dominoes[j], dominoes[k]))
                continue;
                count++;
            }
        }
    }cout<<count;
    return 0;
}