问题描述
代码中的打印函数打印原始电路板而不是解决方案,而求解器函数打印表明原始电路板已更新到位的解决方案。如你所见,我通过引用函数传递了板,那么为什么在调用打印函数后原始板没有得到更新?
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int n = 9;
void print(vector<vector<char>>& board){
for(int i = 0; i < 9; i++){
for(int j = 0; j < n; j++){
cout << board[i][j] << " ";
}
cout << endl;
}
}
void solver(int x,int y,vector<vector<char>>& board,map< pair<int,int>,map<int,int> >& grid,vector<map<int,int>> row,int>>& col){
if(x == 9){
for(int i = 0; i < 9; i++){
for(int j = 0; j < n; j++){
cout << board[i][j] << " ";
}
cout << endl;
}
return;
}
if(y == 9){
solver(x + 1,board,grid,row,col);
return;
}
if(board[x][y] != '.'){
solver(x,y + 1,col);
return;
}
for(int i = 1; i <= 9; i++){
if(!grid[{x/3,y/3}][i] && !row[x][i] && !col[y][i]){
board[x][y] = i + '0';
grid[{x/3,y/3}][i] = 1;
row[x][i] = 1;
col[y][i] = 1;
solver(x,col);
board[x][y] = '.';
grid[{x/3,y/3}][i] = 0;
row[x][i] = 0;
col[y][i] = 0;
}
}
}
void solveSudoku(vector<vector<char>>& board) {
//int n = board.size();
vector< map<int,int> > row(n);
vector< map<int,int> > col(n);
map< pair<int,int> > grid;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(board[i][j] != '.'){
row[i][board[i][j] - '0'] = 1;
col[j][board[i][j] - '0'] = 1;
grid[{i/3,j/3}][board[i][j] - '0'] = 1;
}
}
}
solver(0,col);
}
int main(){
vector<vector<char>> board =
{{ '5','3','.','7','.'},{'6','1','9','5',{'.','8','6',{'8','3'},{'4','1'},{'7','2','6'},'4','5'},'9'}};
solveSudoku(board);
cout << endl;
print(board);
return 0;
}
解决方法
对 solver
的第一次调用是:
solver(0,board,grid,row,col);
因为 board[0][0]
不是 '.'
,所以第一次调用只是
if(board[x][y] != '.'){
solver(x,y + 1,col);
return;
}
即:它调用 solver(0,1,col)
。然后 board[0][1]
是 '.'
,而 x
不是 9
并且 y
不是 9
并且该调用执行:
for(int i = 1; i <= 9; i++){
if(!grid[{x/3,y/3}][i] && !row[x][i] && !col[y][i]){
board[x][y] = i + '0';
grid[{x/3,y/3}][i] = 1;
row[x][i] = 1;
col[y][i] = 1;
solver(x,col);
board[x][y] = '.';
grid[{x/3,y/3}][i] = 0;
row[x][i] = 0;
col[y][i] = 0;
}
}
因此,如果我们内联前两个调用,我们可以用上面的替换 solver(0,col);
得到:
void solveSudoku(vector<vector<char>>& board) {
//int n = board.size();
vector< map<int,int> > row(n);
vector< map<int,int> > col(n);
map< pair<int,int>,map<int,int> > grid;
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(board[i][j] != '.'){
row[i][board[i][j] - '0'] = 1;
col[j][board[i][j] - '0'] = 1;
grid[{i/3,j/3}][board[i][j] - '0'] = 1;
}
}
}
size_t x = 0;
size_t y = 1;
for(int i = 1; i <= 9; i++){
if(!grid[{x/3,y/3}][i] && !row[x][i] && !col[y][i]){
board[x][y] = i + '0';
grid[{x/3,y/3}][i] = 1;
row[x][i] = 1;
col[y][i] = 1;
solver(x,col);
board[x][y] = '.';
grid[{x/3,y/3}][i] = 0;
row[x][i] = 0;
col[y][i] = 0;
}
}
}
这里 board[0][1]
被分配了一个 i + '0';
,然后递归发生,之后 board[0][1]
被重置为 '.'
。当深入递归调用链时,可以应用相同的推理方式。每当您将某些内容分配给 board[x][y]
时,它都会在 solver
返回到 solveSudoku
之前重置。当您处于“递归底部”时,您只能到达 x == 9
并打印更新的板,但您只能通过执行 sudokuSolver
的路径返回 board[x][y] = '.';
。
不知道如何更好地解释,也许使用调试器现场观看它会更有说服力。