这里的第一篇文章，如果我做错了什么，请原谅我。我无法弄清楚如何递归地找到并打印从A点到B点的所有最短路径。
我正在制作一个程序，该程序使用尽可能短的路径使机器人从A点移动到B点。该程序必须使用递归并找到并打印机器人从点A到点B的所有最短路径。例如，如果机器人从点（1、2）开始并且需要以某种方式指向点（3，5），那么程序应该仅用5步就打印出10条路径（例如NNNEE）。这是两个最重要的功能，对不起，代码太乱了。
void Robot::FindPath(const Board &board)
{
string path = "";
vector < string > pathsVector;
MoveRobot(startingX,
startingY,
treasureX,
treasureY,
path,
"",
0,
pathsVector);
}
void Robot::MoveRobot(const int &startingX,
const int &startingY,
const int &treasureX,
const int &treasureY,
const string &path,
const string &lastMove,
const int &numMovesDirection,
vector<string> &pathVector)
{
if (startingX == treasureX && startingY == treasureY)
{
cout << path << endl;
PathCount++;
pathVector.push_back(path);
return;
}
else if (startingY == treasureY)
{
if (startingX - treasureX > 0)
{
if (lastMove == "W" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "W" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX - 1,
startingY,
treasureX,
treasureY,
path + "W",
"W",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "W")
{
MoveRobot(startingX - 1,
startingY,
treasureX,
treasureY,
path + "W",
"W",
1,
pathVector);
}
}
else if (startingX - treasureX < 0)
{
if (lastMove == "E" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "E" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX + 1,
startingY,
treasureX,
treasureY,
path + "E",
"E",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "E")
{
MoveRobot(startingX + 1,
startingY,
treasureX,
treasureY,
path + "E",
"E",
1,
pathVector);
}
}
}
else if (startingX == treasureX)
{
if (startingY - treasureY > 0)
{
if (lastMove == "S" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "S" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX,
startingY - 1,
treasureX,
treasureY,
path + "S",
"S",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "S")
{
MoveRobot(startingX,
startingY - 1,
treasureX,
treasureY,
path + "S",
"S",
1,
pathVector);
}
}
else if (startingY + treasureY > 0)
{
if (lastMove == "N" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "N" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX,
startingY + 1,
treasureX,
treasureY,
path + "N",
"N",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "N")
{
MoveRobot(startingX,
startingY + 1,
treasureX,
treasureY,
path + "N",
"N",
1,
pathVector);
}
}
}
else
{
if (startingX - treasureX > 0)
{
if (lastMove == "W" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "W" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX - 1,
startingY,
treasureX,
treasureY,
path + "W",
"W",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "W")
{
MoveRobot(startingX - 1,
startingY,
treasureX,
treasureY,
path + "W",
"W",
1,
pathVector);
}
}
else if (startingX - treasureX < 0)
{
if (lastMove == "E" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "E" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX + 1,
startingY,
treasureX,
treasureY,
path + "E",
"E",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "E")
{
MoveRobot(startingX + 1,
startingY,
treasureX,
treasureY,
path + "E",
"E",
1,
pathVector);
}
}
else if (startingY - treasureY > 0)
{
if (lastMove == "S" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "S" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX,
startingY - 1,
treasureX,
treasureY,
path + "S",
"S",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "S")
{
MoveRobot(startingX,
startingY - 1,
treasureX,
treasureY,
path + "S",
"S",
1,
pathVector);
}
}
else if (startingY + treasureY > 0)
{
if (lastMove == "N" && numMovesDirection == MaxStepsDir)
{
}
else if (lastMove == "N" &&
(numMovesDirection < MaxStepsDir ||
numMovesDirection > MaxStepsDir))
{
MoveRobot(startingX,
startingY + 1,
treasureX,
treasureY,
path + "N",
"N",
numMovesDirection + 1,
pathVector);
}
else if (lastMove != "N")
{
MoveRobot(startingX,
startingY + 1,
treasureX,
treasureY,
path + "N",
"N",
1,
pathVector);
}
}
}
}
我知道第一个if语句中的“ return”将停止递归，但是当我找到示例中给出的所有10条最短路径并找到并打印时，我不知道在哪里结束递归。因为它只打印EENNN。请忽略if语句，它说“ if（lastMove ==“ W” && numMovesDirection == MaxStepsDir）”，我计划稍后填充这些语句，以使程序的另一部分起作用。任何帮助，我们将不胜感激！
答案 0 :(得分：0)
自己弄清楚。抱歉，如果我在此浪费任何时间。我既有对递归的基本误解，又有代码中的小错误。我没有意识到递归会通过回溯并沿着其他路径自动打印所有最短的路径。我的代码中阻止此错误的错误是，在代码末尾的大“ else”中，我无意中输入了“ else if（startingY-宝物Y> 0）”。这应该只是说“ if（startY-宝物Y> 0）”，因为如果没有一个合适的位置，那么大的else代码块的全部要点就是改变x坐标和y坐标，如果没有则将其放到其他位置那。一旦将else if更改为if，它将打印出我给出的示例的所有十个最短路径。