Dominosa/数邻（2） | C++ | BFS

一、Dominosa简介

   Dominosa，中文名称为数邻，是一种棋盘游戏，基于骨牌的排列和匹配来进行。它是从骨牌游戏中发展而来的，在骨牌的基础上添加了一些规则和难度。具体的游戏规则是：将一副骨牌放置在一个棋盘上，玩家需要根据这些固定数字推断出正确的骨牌排列。

   Dominosa 是一款非常有趣和具有挑战性的游戏。我是在大三小学期的 Linux 实训中，在虚拟机上接触到了这个小游戏。由于我本身玩过骨牌，同时对这种类型的数学游戏也很感兴趣，它们实在是令我着迷。

   Dominosa 需要玩家运用逻辑思维和推理能力来解决问题。它的规则相对简单，但难度却很高，因为骨牌排列的可能性非常多。如果您喜欢逻辑游戏，并且想尝试一些挑战性的新游戏，那么 Dominosa 绝对值得一试！

   对于没有接触过骨牌的同学来说或许还不太能理解这个游戏，那么我们下面举一个具体的例子。下面是一个经典的 Dominosa 谜面。

答案如下：

   可见，盘面上的数字被两两划分，且保证每个划分都互不重复。

   我相信看完例子你一定懂了这个游戏的规则！现在，请你编写一个程序，对于给定的 Dominosa 谜面，能够自动判断是否有解，如果有解，还能够给出一个正确的解。

二、题目描述

输入一个未解的 Dominosa 谜面，输出它的结果。
Input
由两部分输入组成。第一部分只有一个正整数n，表示最大数字，第二部分有n+1行，每行n+2个正整数，表示谜面。
Output
谜题的解，如果两个数字能够组成一个骨牌，那么将它们的位置标记为相同的数字。标记方法可能不唯一，但是标记必须容易辨认。

测试用例1

测试用例2

输入（供复制）：
9
82419008183
64907897150
03198511218
68804535022
37763052729
74416693226
36165567745
81442980856
42513904405
93269977337

三、编程思路

在昨天的代码基础上，

1. 将求解的步骤单独写成一个函数
2. 当需要用到BFS的时候，可以递归调用这个求解函数
3. 当找到正确解后，输出正确解，然后逐层返回，程序结束

四、完整代码

说明：代码中默认为输出到命令行中，我们也提供了输出到文件中的选项，只要在程序的第 390-395 行，更改被注释的内容即可。

#include
using namespace std;
int N = 30;							// N: 最大数字，不难计算出盘面的行数=N+1，列数=N+2
// 定义单格
typedef struct {
	int number = -1;						// 表示当前格内的数字，初始是-1
	int con = 0;							// 表示当前格是否已建立连接，如果未连接，则为0，如果已连接，则为1
	int up=1, down=1, left=1, right=1;		// 标记当前格的上、下、左、右格子是否可能建立连接，如果可能则为1，如果不能则为0
	int sequ = 0;
} Single;

// 定义骨牌
typedef struct {
	int count;				// 编号：以6为例，00->0, 01->1, ..., 06->6, 11->7, ..., 16->12, 22->13, ..., 55->25, 56->26, 66->27；count是每种骨牌的数量
	bool used;				// used是骨牌是否已经出现过，若已经出现过则为true
} bridge;

int addd(int m){			// 一个求和的函数
	int s = 0;
	for (int i=1; i<m; i++) s+=i;
	return s;
}
// 根据骨牌上的两个数字将其转换为对应的b_id
int trans_numbers_bid(int number1, int number2) {	
	int addd(int);		
	int less = number1 < number2 ? number1 : number2, more = number1 + number2 - less;
	int b_id = less*N+more;
	if (less>1) b_id-=addd(less);
	return b_id;
}

// Dominosa棋盘类
class Dominosa{
public:
	Dominosa(){};				// 构造函数
	Dominosa(const Dominosa&);	// 拷贝函数
	~Dominosa(){};				// 析构函数
	Single lattice[31][32];		// 第i行第j列的格子位置为：k=21*i+j，支持的最大数字 N=30
	bridge b[31*32/2];
	void init_connect();		// 初始化所有格子的up、down、left、right
	int sum_of_up_down_left_right(int m, int n); // 计算上下左右的和
	int sequence = 1;
};
Dominosa:: Dominosa(const Dominosa& dd) {
	for (int i=0; i<21; i++) {
		for (int j=0; j<22; j++) {
			this->lattice[i][j] = dd.lattice[i][j];
		}
	}
	for (int i=0; i<(N+1)*(N+2)/2; i++) {
		this->b[i].used = dd.b[i].used;
	}
	this -> sequence = dd.sequence;
}
void Dominosa:: init_connect(){
	// 初始化所有格子的up、down、left、right
	for (int j=0; j<N+2; j++) {
		lattice[0][j].up = 0;
		lattice[N][j].down = 0;
	}
	for (int i=0; i<N+1; i++) {
		lattice[i][0].left = 0;
		lattice[i][N+1].right = 0;
	}
	// 初始化 bridge
	for (int i=0; i<(N+1)*(N+2)/2; i++) {
		b[i].used = false;
	}
}
// 计算当前格与上下左右可能性的和，m是行数，n是列数
int Dominosa:: sum_of_up_down_left_right(int m, int n){
	// 如果是边缘格子，单独考虑
	if (m==0) {
		if (n==0) {
			return lattice[m][n].right+lattice[m][n].down;
		}
		else if (n==N+1) {
			return lattice[m][n].left+lattice[m][n].down;
		}
		else {
			return lattice[m][n].left+lattice[m][n].right+lattice[m][n].down;
		}
	}
	else if (m==N) {
		if (n==0) {
			return lattice[m][n].up+lattice[m][n].right;
		}
		else if (n==N+1) {
			return lattice[m][n].left+lattice[m][n].up;
		}
		else {
			return lattice[m][n].left+lattice[m][n].right+lattice[m][n].up;
		}
	}
	else {
		if (n==0) {
			return lattice[m][n].right+lattice[m][n].up+lattice[m][n].down;
		}
		else if (n==N+1) {
			return lattice[m][n].left+lattice[m][n].up+lattice[m][n].down;
		}
		else {
			return lattice[m][n].up+lattice[m][n].down+lattice[m][n].left+lattice[m][n].right;
		}
	}
}

// 求解 Dominosa 谜面，并调用print函数输出解
int D_solve(Dominosa D) {		// 返回-1，表示无解；返回0，表示可能有解，但需要BFS进一步计算；返回1，表示成功解出。

	void print_solution(Dominosa);
	void print_solution_in_file(Dominosa);

	int count_of_bridges = (N+1)*(N+2)/2;
	bool update = true;
	while (update) {
		update = false;
		
		// 初始化bridge.count
		for (int i=0; i<count_of_bridges; i++) {
			D.b[i].count = 0;
		}
		// 全面扫描D，更新bridge.count
		// 先横向扫描一遍，更新bridge.count
		for (int i=0; i<N+1; i++) {
			for (int j=0; j<N+1; j++) {
				if (D.lattice[i][j].con==0 && D.lattice[i][j+1].con==0) {
					D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].count++;
				}
			}
		}
		// 再纵向扫描一遍，更新bridge.count
		for (int i=0; i<N; i++) {
			for (int j=0; j<N+2; j++) {
				if (D.lattice[i][j].con==0 && D.lattice[i+1][j].con==0) {
					D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].count++;
				}
			}
		}
		// 全面扫描一遍，检查是否有四面都连接不了的格子，如果有，直接返回-1
		for (int i=0; i<N+1; i++) {
			for (int j=0; j<N+2; j++) {
				if (D.lattice[i][j].con==0 && D.sum_of_up_down_left_right(i, j)==0) {
					return -1;
				}
			}
		}

		// 从左向右扫描一遍，并尝试向右建立连接
		for (int i=0; i<N+1; i++) {
			for (int j=0; j<N+1; j++) {
				if (D.lattice[i][j].con==0 && D.lattice[i][j+1].con==0) {
					if (D.sum_of_up_down_left_right(i, j)==1 && D.lattice[i][j].right==1) {
						D.lattice[i][j].con = 1;
						D.lattice[i][j].sequ = D.sequence;
						D.lattice[i][j+1].con = 1;
						D.lattice[i][j+1].sequ = D.sequence;
						D.sequence++;
						D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used = true;
						update = true;
					}
					if (D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used==false
						&& D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].count==1) {
							D.lattice[i][j].con = 1;
							D.lattice[i][j].sequ = D.sequence;
							D.lattice[i][j+1].con = 1;
							D.lattice[i][j+1].sequ = D.sequence;
							D.sequence++;
							D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used = true;
							update = true;
					}
				}
			}
		}
		// 从右向左扫描一遍，并尝试向左建立连接
		for (int i=0; i<N+1; i++) {
			for (int j=1; j<N+2; j++) {
				if (D.lattice[i][j].con==0 && D.lattice[i][j-1].con==0) {
					if (D.sum_of_up_down_left_right(i, j)==1 && D.lattice[i][j].left==1) {
						D.lattice[i][j].con = 1;
						D.lattice[i][j].sequ = D.sequence;
						D.lattice[i][j-1].con = 1;
						D.lattice[i][j-1].sequ = D.sequence;
						D.sequence++;
						D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used = true;
						update = true;
					}
					if (D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used==false
						&& D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].count==1) {
							D.lattice[i][j].con = 1;
							D.lattice[i][j].sequ = D.sequence;
							D.lattice[i][j-1].con = 1;
							D.lattice[i][j-1].sequ = D.sequence;
							D.sequence++;
							D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used = true;
							update = true;
					}
				}
			}
		}
		// 从上向下扫描一遍，并尝试向下建立连接
		for (int i=0; i<N; i++) {
			for (int j=0; j<N+2; j++) {
				if (D.lattice[i][j].con==0 && D.lattice[i+1][j].con==0) {
					if (D.sum_of_up_down_left_right(i, j)==1 && D.lattice[i][j].down==1) {
						D.lattice[i][j].con = 1;
						D.lattice[i][j].sequ = D.sequence;
						D.lattice[i+1][j].con = 1;
						D.lattice[i+1][j].sequ = D.sequence;
						D.sequence++;
						D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used = true;
						update = true;
					}
					if (D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used==false
						&& D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].count==1) {
							D.lattice[i][j].con = 1;
							D.lattice[i][j].sequ = D.sequence;
							D.lattice[i+1][j].con = 1;
							D.lattice[i+1][j].sequ = D.sequence;
							D.sequence++;
							D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used = true;
							update = true;
					}
				}
			}
		}
		// 从下向上扫描一遍，并尝试向上建立连接
		for (int i=1; i<N+1; i++) {
			for (int j=0; j<N+2; j++) {
				if (D.lattice[i][j].con==0 && D.lattice[i-1][j].con==0) {
					if (D.sum_of_up_down_left_right(i, j)==1 && D.lattice[i][j].up==1) {
						D.lattice[i][j].con = 1;
						D.lattice[i][j].sequ = D.sequence;
						D.lattice[i-1][j].con = 1;
						D.lattice[i-1][j].sequ = D.sequence;
						D.sequence++;
						D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used = true;
						update = true;
					}
					if (D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used==false
						&& D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].count==1) {
							D.lattice[i][j].con = 1;
							D.lattice[i][j].sequ = D.sequence;
							D.lattice[i-1][j].con = 1;
							D.lattice[i-1][j].sequ = D.sequence;
							D.sequence++;
							D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used = true;
							update = true;
					}
				}
			}
		}

		// 全面更新D的每个con=0格子的up、down、left、right状态
		// 如果相邻的格子的con=1，则将其方向设置为0
		// 如果相邻的格子的con=0，但bridge.used=true，也要将其方向设置为0
		{
			int i=0;
			{
				int j=0;
				if (D.lattice[i][j].con==0) {
					if (D.lattice[i][j+1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used==true ) {
						D.lattice[i][j].right = 0;
					}
					if (D.lattice[i+1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used==true ) {
						D.lattice[i][j].down = 0;
					}
				}
				for (j=1; j<N+1; j++) {
					if (D.lattice[i][j].con==0) {
						if (D.lattice[i][j-1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used==true ) {
							D.lattice[i][j].left = 0;
						}
						if (D.lattice[i][j+1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used==true ) {
							D.lattice[i][j].right = 0;
						}
						if (D.lattice[i+1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used==true ) {
							D.lattice[i][j].down = 0;
						}
					}
				}
				j = N+1;
				if (D.lattice[i][j].con==0) {
					if (D.lattice[i][j-1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used==true ) {
						D.lattice[i][j].left = 0;
					}
					if (D.lattice[i+1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used==true ) {
						D.lattice[i][j].down = 0;
					}
				}
			}

			{
				for (i=1; i<N; i++) {
					int j=0;
					if (D.lattice[i][j].con==0) {
						if (D.lattice[i][j+1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used==true ) {
							D.lattice[i][j].right = 0;
						}
						if (D.lattice[i-1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used==true ) {
							D.lattice[i][j].up = 0;
						}
						if (D.lattice[i+1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used==true ) {
							D.lattice[i][j].down = 0;
						}
					}
					for (j=1; j<N+1; j++) {
						if (D.lattice[i][j].con==0) {
							if (D.lattice[i][j-1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used==true ) {
								D.lattice[i][j].left = 0;
							}
							if (D.lattice[i][j+1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used==true ) {
								D.lattice[i][j].right = 0;
							}
							if (D.lattice[i-1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used==true ) {
								D.lattice[i][j].up = 0;
							}
							if (D.lattice[i+1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used==true ) {
								D.lattice[i][j].down = 0;
							}
						}
					}
					j = N+1;
					if (D.lattice[i][j].con==0) {
						if (D.lattice[i][j-1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used==true ) {
							D.lattice[i][j].left = 0;
						}
						if (D.lattice[i-1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used==true ) {
							D.lattice[i][j].up = 0;
						}
						if (D.lattice[i+1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i+1][j].number)].used==true ) {
							D.lattice[i][j].down = 0;
						}
					}
				}
			}

			{
				i = N;
				int j=0;
				if (D.lattice[i][j].con==0) {
					if (D.lattice[i][j+1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used==true ) {
						D.lattice[i][j].right = 0;
					}
					if (D.lattice[i-1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used==true ) {
						D.lattice[i][j].up = 0;
					}
				}
				for (j=1; j<N+1; j++) {
					if (D.lattice[i][j].con==0) {
						if (D.lattice[i][j-1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used==true ) {
							D.lattice[i][j].left = 0;
						}
						if (D.lattice[i][j+1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j+1].number)].used==true ) {
							D.lattice[i][j].right = 0;
						}
						if (D.lattice[i-1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used==true ) {
							D.lattice[i][j].up = 0;
						}
					}
				}
				j = N+1;
				if (D.lattice[i][j].con==0) {
					if (D.lattice[i][j-1].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i][j-1].number)].used==true ) {
						D.lattice[i][j].left = 0;
					}
					if (D.lattice[i-1][j].con==1 || D.b[trans_numbers_bid(D.lattice[i][j].number, D.lattice[i-1][j].number)].used==true ) {
						D.lattice[i][j].up = 0;
					}
				}
			}
		}
	}

	int fbi=-1, fbj=-1;
	// 检测是否已经成功解出，如果未解出，标记第一个未连接的格子
	bool if_solve = true;
	for (int i=0; i<N+1; i++) {
		for (int j=0; j<N+2; j++) {
			if (D.lattice[i][j].con==0) {
				fbi = i;
				fbj = j;
				if_solve = false;
				break;
			}
		}
		if (if_solve==false) break;
	}

	// 如果已经解出，则返回true
	if (if_solve==true) {
		print_solution(D);	// 输出到命令行中
		// print_solution_in_file(D);	// 输出到文件 output.txt 中
		return 1;
	}
	
	// 否则用BFS求解(其实用的是递归啦哈哈)
	Dominosa d1(D), d2(D);
	// 设置d1为向右连接
	d1.lattice[fbi][fbj  ].con = 1;	d1.lattice[fbi][fbj  ].sequ = d1.sequence;
	d1.lattice[fbi][fbj+1].con = 1;	d1.lattice[fbi][fbj+1].sequ = d1.sequence;
	d1.sequence++;
	d1.b[trans_numbers_bid(d1.lattice[fbi][fbj].number, d1.lattice[fbi][fbj+1].number)].used = true;
	int solve_d1 = D_solve(d1);
	if (solve_d1==1) {
		return 1;
	}

	// 设置d2为向下连接
	d2.lattice[fbi  ][fbj].con = 1; d2.lattice[fbi  ][fbj].sequ = d2.sequence;
	d2.lattice[fbi+1][fbj].con = 1; d2.lattice[fbi+1][fbj].sequ = d2.sequence;
	d2.sequence++;
	d2.b[trans_numbers_bid(d2.lattice[fbi][fbj].number, d2.lattice[fbi+1][fbj].number)].used = true;
	int solve_d2 = D_solve(d2);
	if (solve_d2==1) {
		return 1;
	}

	return 0;
}

// 在命令行中输出最终的结果
void print_solution(Dominosa D) {
	for (int i=0; i<N+1; i++) {
		for (int j=0; j<N+2; j++) {
			cout << setw(4) << D.lattice[i][j].sequ;
		}
		cout << endl;
	}
}

// 将结果输出到 output.txt 中
void print_solution_in_file(Dominosa D) {
	fstream f;
	f.open("output.txt", ios::out|ios::app);
	f << endl; // 如果output.txt中本身已经有内容了，可以将其区分开，避免混淆

	for (int i=0; i<N+1; i++) {
		for (int j=0; j<N+2; j++) {
			f << setw(4) << D.lattice[i][j].sequ;
		}
		f << endl;
	}
	f.close();
}

Dominosa DMNS;				// 基础盘面DMNS
int main() {
	
	int trans_numbers_bid(int, int);
	int D_solve(Dominosa);

	// 初始化谜面
	DMNS.init_connect();

    freopen("input.txt","r",stdin);		// 谜面放在 input.txt 中
    //读入待解盘面D
	scanf("%d\n", &N);
	char c;
	for (int i=0; i<N+1; i++) {
		for (int j=0; j<N+2; j++) {
			scanf("%c", &c);
			DMNS.lattice[i][j].number = c - '0';
			DMNS.lattice[i][j].con = 0;
		}
		scanf("%*c");
	}
	
	int r = D_solve(DMNS);
	
    return 0;
}

现在的代码应该能够做到求解出每一个 Dominosa 谜题的答案，当然，你得确保谜题本身有解。