递归函数在0x79B20AD2(ucrtbased.dll)处引发未处理的异常

Question

我编写了minesweeper，目前的任务是编写一个函数，以发现附近没有地雷的区域。在最初的扫雷舰中，如果你在没有地雷的区域内点击，它会打开一个区域，直到边界附近有地雷。为此，我编写了函数dissolve((。这是代码：

#include <iostream>
#include <cstdlib>
#include <time.h>
#include <vector>
#include <string>
using namespace std;
#define Str1D vector<string>
#define Str2D vector<Str1D>
#define Int1D vector<int>
#define Int2D vector<Int1D>
void unravel(Str2D &fogofwar, Int2D &display, int x, int y) {
for (int minusrows = -1; minusrows < 2; minusrows++){ // going through the 
// neighbouring cells (+ the cell itself)
for (int minuscolumns = -1; minuscolumns < 2; minuscolumns++){
if (x + minusrows > 0 && y + minuscolumns > 0 && x + minusrows < fogofwar.size() && y + minuscolumns < fogofwar[0].size()){ // checking 
                                                                  // if within borders
if (x > 0 && y > 0 && x < fogofwar.size() && y < fogofwar[0].size()) { // checking if the oririginal 
                 // values are within borders
fogofwar[x + minusrows][y + minuscolumns] = to_string(display[x + minusrows][y + minuscolumns]); // revealing the 
                                               // neighbouring cells
if (display[x + minusrows][y + minuscolumns] == 0) { // if the cell is 0 on the display, 
   // open it and the 8 neighbouring to it cells
if (not (minusrows == 0 && minuscolumns == 0)) { // if it's not the same cell, of course,
   // otherwise it's an endless cycle
unravel(fogofwar, display, x + minusrows, y + minuscolumns);
}
}
}                               
}
}
}
}

int main() {
int row, column, prob;
bool running = true;

cout << "Input width and height: ";
cin >> row >> column;
cout << endl << "Input mines probability (%): ";
cin >> prob;
cout << endl;
srand (time(NULL));
Int2D field(row + 1, Int1D(column + 1));
Int2D display(row + 1, Int1D(column + 1));
Str2D fogofwar(row + 1, Str1D(column + 1, "*"));
field[0][0] = 0; // field of mines
display[0][0] = 0; // display of neighbouring mines
fogofwar[0][0] = to_string(0); // what the player will see
for (int i = 1; i < row + 1; i++) { //assigning coordinates
field[i][0] = i;
display[i][0] = i;
fogofwar[i][0] = to_string(i);
}
for (int j = 1; j < column + 1; j++) { //assigning coordinates
field[0][j] = j;
display[0][j] = j;
fogofwar[0][j] = to_string(j);
}
for (int i = 1; i < row + 1; i++){ // filling the field with mines
for (int j = 1; j < column + 1; j++){
int x = rand() % 100;
if (x < prob) {
field[i][j] = 1;
}
else{
field[i][j] = 0;
}
}
}
cout << endl << endl;
for (int i = 0; i < row + 1; i++){ // printing field
for (int j = 0; j < column + 1; j++){
cout << " " << field[i][j] << " ";
}
cout << endl;
}
cout << endl << endl;
for (int i = 0; i < row + 1; i++){ // assigning the display of amount of neighbouring mines
for (int j = 0; j < column + 1; j++){
int count = 0;
if (i > 0 && j > 0){
for (int minusrows = -1; minusrows < 2; minusrows++){
for (int minuscolumns = -1; minuscolumns < 2; minuscolumns++){
if (i + minusrows > 0 && i + minusrows < row + 1 && j + minuscolumns > 0 && j + minuscolumns < column + 1){
if (field[i + minusrows][j + minuscolumns] == 1){
count++;
}
}
}
}
display[i][j] = count;
}

cout << " " << display[i][j] << " ";

}
cout << endl;
}

cout << endl << endl;

while (running) {
for (int i = 0; i < row + 1; i++){
for (int j = 0; j < column + 1; j++){
cout << " " << fogofwar[i][j] << " ";
}
cout << endl;
}
cout << endl;
int x, y;
cout << endl << "Input the target cell (x, y): ";
cin >> x >> y;
cout << endl;
unravel(fogofwar, display, x, y);
}
return 0;
}

如果我通过更改unlock(fogofwar，display，x+minusrows，y+minuscolumns(来删除递归性到继续在函数dissolve((中，它按预期工作。但我需要打开显示屏上有0的整个区域。有没有办法绕过这个错误或永远纠正它？

Answer 1

首先，我无法用有问题的信息重现错误。请尝试指定完整的用例以及出现错误的值。但是，在实施中存在一个明显的问题。你多次访问同一个单元格，直到内存超过总内存(我相信这就是你的程序崩溃的原因(

您应该保留已访问的插槽。你可以通过多种方式做到这一点。我正在提供一种处理此问题的方法。

尝试以下代码：-

#include <iostream>
#include <cstdlib>
#include <time.h>
#include <vector>
#include <string>
using namespace std;
#define Str1D vector<string>
#define Str2D vector<Str1D>
#define Int1D vector<int>
#define Int2D vector<Int1D>
void unravel(Str2D &fogofwar, Int2D &display, int x, int y, vector<vector<bool> > &visited) {
for (int minusrows = -1; minusrows < 2; minusrows++){ // going through the
// neighbouring cells (+ the cell itself)
for (int minuscolumns = -1; minuscolumns < 2; minuscolumns++){
if (x + minusrows > 0 && y + minuscolumns > 0 && x + minusrows < fogofwar.size() && y + minuscolumns < fogofwar[0].size()){ // checking
                                                                  // if within borders
if (x > 0 && y > 0 && x < fogofwar.size() && y < fogofwar[0].size()) { // checking if the oririginal
                 // values are within borders
if (x > 0 && y > 0 && x < visited.size() && y < visited[0].size()) {
cout.flush();
}
visited[x][y] = true;
fogofwar[x + minusrows][y + minuscolumns] = to_string(display[x + minusrows][y + minuscolumns]); // revealing the
                                               // neighbouring cells
if (display[x + minusrows][y + minuscolumns] == 0) { // if the cell is 0 on the display,
   // open it and the 8 neighbouring to it cells
if (not visited[x + minusrows][y + minuscolumns]) { // if it's not the same cell, of course,
   // otherwise it's an endless cycle
unravel(fogofwar, display, x + minusrows, y + minuscolumns, visited);
}
}
}
}
}
}
}

int main() {
int row, column, prob;
bool running = true;

cout << "Input width and height: ";
cin >> row >> column;
cout << endl << "Input mines probability (%): ";
cin >> prob;
cout << endl;
srand (time(NULL));
Int2D field(row + 1, Int1D(column + 1));
Int2D display(row + 1, Int1D(column + 1));
Str2D fogofwar(row + 1, Str1D(column + 1, "*"));
field[0][0] = 0; // field of mines
display[0][0] = 0; // display of neighbouring mines
fogofwar[0][0] = to_string(0); // what the player will see
for (int i = 1; i < row + 1; i++) { //assigning coordinates
field[i][0] = i;
display[i][0] = i;
fogofwar[i][0] = to_string(i);
}
for (int j = 1; j < column + 1; j++) { //assigning coordinates
field[0][j] = j;
display[0][j] = j;
fogofwar[0][j] = to_string(j);
}
for (int i = 1; i < row + 1; i++){ // filling the field with mines
for (int j = 1; j < column + 1; j++){
int x = rand() % 100;
if (x < prob) {
field[i][j] = 1;
}
else{
field[i][j] = 0;
}
}
}
cout << endl << endl;
for (int i = 0; i < row + 1; i++){ // printing field
for (int j = 0; j < column + 1; j++){
cout << " " << field[i][j] << " ";
}
cout << endl;
}
cout << endl << endl;
for (int i = 0; i < row + 1; i++){ // assigning the display of amount of neighbouring mines
for (int j = 0; j < column + 1; j++){
int count = 0;
if (i > 0 && j > 0){
for (int minusrows = -1; minusrows < 2; minusrows++){
for (int minuscolumns = -1; minuscolumns < 2; minuscolumns++){
if (i + minusrows > 0 && i + minusrows < row + 1 && j + minuscolumns > 0 && j + minuscolumns < column + 1){
if (field[i + minusrows][j + minuscolumns] == 1){
count++;
}
}
}
}
display[i][j] = count;
}

cout << " " << display[i][j] << " ";

}
cout << endl;
}

cout << endl << endl;

while (running) {
for (int i = 0; i < row + 1; i++){
for (int j = 0; j < column + 1; j++){
cout << " " << fogofwar[i][j] << " ";
}
cout << endl;
}
cout << endl;
int x, y;
cout << endl << "Input the target cell (x, y): ";
cin >> x >> y;
cout << endl;
vector<vector<bool> > visited(row+1, vector<bool>(column+1, false));
unravel(fogofwar, display, x, y, visited);
}
return 0;
}

变化是，我保持着一个参观过的阵列，我再也不会回到我以前在解开中去过的地方。