目录
- 一、Dominosa简介
- 二、题目描述
- 三、编程思路
- 四、完整代码
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，
- 将求解的步骤单独写成一个函数
- 当需要用到BFS的时候，可以递归调用这个求解函数
- 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<bits/stdc++.h>
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。
