我正在寻找一些有关我之前曾隐约询问过的问题的帮助,该问题以递归方式解决了15钉纸牌。编译和运行它时,我总是收到奇怪的错误,其中大多数都说“堆栈溢出”,或者我遇到了段错误。这就是我到目前为止的内容,其中“ board [15]”代表15钉板,“ moves [36]”代表可以进行的所有可能动作。当只剩下一个钉时,应该发现递归。
#include <iostream>
using namespace std;
void solveGame(int a[15], int b[36][3], int c[15][4]);
void chooseMove (int a[15], int b[36][3], int openSpace, int c[15][4]);
int findEmpty (int a[15]);
int pegCount (int a[15]);
bool isPeg (int peg, int a[15]);
int usedVals[15] = {-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1,-1};
int d = 0;
int index = 0;
int main ()
{
int openSpace = 5;
int board[15]= {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
board[openSpace] = 0;
int alreadyMoved[15][4];
int moves[36][3] = {{0, 1, 3},
{0, 2, 5},
{1, 3, 6},
{1, 4, 8},
{2, 4, 7},
{2, 5, 9},
{3, 6, 10},
{3, 7, 12},
{3, 1, 0},
{3, 4, 5},
{4, 7, 11},
{4, 8, 13},
{5, 9, 14},
{5, 8, 12},
{5, 2, 0},
{5, 4, 3},
{6, 3, 1},
{6, 7, 8},
{7, 4, 2},
{7, 8, 9},
{8, 4, 1},
{8, 7, 6},
{9, 5, 2},
{9, 8, 7},
{10, 6, 3},
{10, 11, 12},
{11, 7, 4},
{11, 12, 13},
{12, 7, 3},
{12, 8, 5},
{12, 11, 10},
{12, 13, 14},
{13, 8, 4},
{13, 12, 11},
{14, 9, 5},
{14, 13, 12}};
solveGame(board, moves, alreadyMoved);
for (int i = 0; i < 13; i++)
cout << alreadyMoved[i][0] << " " << alreadyMoved[i][1] << " " < <alreadyMoved[i][2] << endl;
return 0;
}
// main recursive function
void solveGame (int a[15], int b[36][3], int c[15][4]
{
int empSpace;
int moveIndex;
if (pegCount(a) < 2) {
cout<<"game over"<<endl;
} else {
empSpace = findEmpty(a);
chooseMove(a, b, empSpace, c);
solveGame(a, b, c);
}
}
// supposed to pick a move that is applicable to the board otherwise it find a new move
void chooseMove (int a[15], int b[36][3], int openSpace, int c[15][4])
{
int i = 0;
while (1) {
if (i < 36 && b[i][2] == openSpace && isPeg(b[i][0],a) && isPeg(b[i][1],a)) {
a[b[i][0]] = 0;
a[b[i][1]] = 0;
a[b[i][2]] = 1;
c[d][0] = b[i][0];
c[d][1] = b[i][1];
c[d][2] = b[i][2];
c[d][3] = i;
d++;
index = 0;
for (int v = 0; v < 15; v++)
usedVals[v] = -1;
break;
} else if (i > 35) {
a[b[c[d-1][3]][0]] = 1;
a[b[c[d-1][3]][1]] = 1;
a[b[c[d-1][3]][2]] = 0;
c[d-1][0] = 0;
c[d-1][1] = 0;
c[d-1][2] = 0;
c[d-1][3] = 0;
usedVals[index] = openSpace;
index++;
int newOpen = findEmpty(a);
chooseMove(a, b, newOpen, c);
}
i++;
}
}
// counts the pegs on the board in order to cancel recursion
int pegCount (int a[15])
{
int count = 0;
for (int i = 0; i < 15; i++)
if (a[i] == 1)
count++;
return count;
}
// finds an empty space that hasn't already been found faulty
int findEmpty (int a[15])
{
for (int i = 0; i < 15; i++) {
for(int j = 0; j < 15; j++) {
if(a[i] == 0 && i != usedVals[j] && usedVals[j] > -1)
return i;
}
}
}
// tests if current index is a peg
bool isPeg (int peg, int a[15])
{
return a[peg] == 1;
}
最佳答案
快速浏览显示了很多潜在的问题,但是我认为这可能归结为您传递数组的方式。数组是按引用而不是按值传递的,因此递归函数正在处理数组的单个副本,我认为这不是您想要的。因此,您永远都找不到最终的举动,这将使您从无限递归中获得堆栈溢出。
尝试在每个递归级别分配新的数组副本。有些人会希望您为此使用new
或malloc
,因为他们认为对C ++的介绍应该是一次尝试,您必须掌握内存管理才能做任何有用的事情。相反,我建议您不要使用任何数组。使用一个集合类,该集合类在按值传递时将正常工作(我认为POD的std :: vector会做到这一点),并且该集合类将按照代码似乎期望的方式创建数组的副本。
当您确实想要广度优先搜索时,您可能还存在在chooseMove中进行深度优先搜索的问题。
关于c++ - 递归解决方案未按预期运行/出现错误,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/9196487/