C++ 网格系统中的寻路
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Pathfinding in a grid system
提问by Sefam
I'm currently building a small grid-based game in C++ using SDL. I've made a tile class, which represents each individual tile on a map. This tile class is used in a 2 dimensional vector, one dimension representing the X axis and the other one the Y axis.
我目前正在使用 SDL 用 C++ 构建一个基于网格的小型游戏。我创建了一个 tile 类，它代表地图上的每个单独的 tile。此 tile 类用于二维向量，一维表示 X 轴，另一维表示 Y 轴。
I'm having an algorithm issue, I do not even know where to start with this, let's say I have this map:
我有一个算法问题，我什至不知道从哪里开始，假设我有这张地图：
0 0 1 1 0 E
0 0 0 1 0 1
0 C 0 1 0 1
0 1 0 1 0 1
0 1 1 1 1 1
C is my character, E is an exit, and 1 is a floor tile.
C是我的角色，E是出口，1是地砖。
I would like to figure out the most optimal way to figure out if there's a way for the character to reach the exit. I know I could use a function to manually check every tile around C, and for every floor tile around C, I check again every tile around, until I find a consistent path to get to E, but that doesn't seem very optimal.
我想找出最佳方法来确定角色是否有办法到达出口。我知道我可以使用一个函数来手动检查 C 周围的每个瓷砖，对于 C 周围的每个地砖，我再次检查周围的每个瓷砖，直到找到到达 E 的一致路径，但这似乎不是最佳选择。
Can I have a clue or some sort of direction in which to orient myself?
我可以有线索或某种方向来定位自己吗？
回答by thefourtheye
There are so many algorithms which can find paths between two points. There are three algorithms which are easy to implement and understand.
有很多算法可以找到两点之间的路径。共有三种易于实现和理解的算法。
- Depth first search (DFS)
- Breadth first search (BFS)
- Dijkstra's algorithm
- 深度优先搜索（DFS）
- 广度优先搜索（BFS）
- Dijkstra 算法
Depth first search
深度优先搜索
This algorithm, takes the current node, finds all the neighbors, puts them in a stack, pops by one and traverses till the end or it finds the path.
该算法采用当前节点，找到所有邻居，将它们放入堆栈中，弹出一个并遍历直到最后或找到路径。
Breadth first search
广度优先搜索
This algorithm, takes the current node, finds all the neighbors, puts them in a queue, dequeues one by one and traverses till the end or it finds the path.
这个算法，取当前节点，找到所有的邻居，把它们放入队列，一个一个地出队，遍历到最后或者找到路径。
The difference between DFS and BFS is that, DFS cannot guarantee optimal solution. Consider this case.
DFS和BFS的区别在于，DFS不能保证最优解。考虑这个案例。
S 1 1
1 1 1
1 1 E
Assuming S is (0,0) and E is (2, 2). There are many optimal solutions to this maze. Since, DFS checks the path of its neighbours till the end, it might take S -> (1,0) -> (2,0) -> (2,1) -> (1,1) -> (1,2) -> Eand it will return 6 as the cost of the path. Whereas, BFS finds all the neighbours, neighbours of all neighbours and goes on. If one of neighbors is E, it returns the cost. That would be guaranteed to be optimal. So, BFS might go like this. S -> (1,0) -> (2,0) -> (2,1) -> E(It finds neighbours of neighbours, it does not go till the end with each of the neighbours).
假设 S 是 (0,0)，E 是 (2, 2)。这个迷宫有很多最优解。因为，DFS 直到最后都会检查其邻居的路径，所以它可能需要S -> (1,0) -> (2,0) -> (2,1) -> (1,1) -> (1,2) -> E并且将返回 6 作为路径的成本。而 BFS 找到所有邻居，所有邻居的邻居并继续。如果邻居之一是 E，则返回成本。这将保证是最佳的。所以，BFS 可能是这样的。S -> (1,0) -> (2,0) -> (2,1) -> E（它找到邻居的邻居，它不会与每个邻居走到最后）。
Dijkstra's algorithm
Dijkstra 算法
It is similar to BFS, but it can have weights. In the example, we assumed that the cost of moving from one node to another costs us 1 unit. In Dijkstra's algorithm, it allows us to use any positive integer as the cost, and it can be different for each and every link.
它类似于 BFS，但它可以有权重。在示例中，我们假设从一个节点移动到另一个节点的成本为 1 个单位。在 Dijkstra 算法中，它允许我们使用任何正整数作为成本，并且每个链接的成本都可以不同。
Conclusion
结论
If you want optimal result, go for BFS or Dijkstra's algorithm. For your case, BFS should work.
如果您想要最佳结果，请使用 BFS 或 Dijkstra 算法。对于您的情况，BFS 应该可以工作。
回答by Pier
Take a look at the most used path finding algorithms.
看看最常用的寻路算法。
http://qiao.github.io/PathFinding.js/visual/
http://qiao.github.io/PathFinding.js/visual/
This is done in JS, but you should be able to find a C++ implementation of the one that fits your needs, or write your own.
这是在 JS 中完成的，但您应该能够找到适合您需求的 C++ 实现，或者自己编写。
回答by Mehul Nirala
#include<bits/stdc++.h>
using namespace std;
#define ROW 9
#define COL 10
// Creating a shortcut for int, int pair type
typedef pair<int, int> Pair;
// Creating a shortcut for pair<int, pair<int, int>> type
typedef pair<double, pair<int, int> > pPair;
// A structure to hold the neccesary parameters
struct cell
{
// Row and Column index of its parent
// Note that 0 <= i <= ROW-1 & 0 <= j <= COL-1
int parent_i, parent_j;
// f = g + h
double f, g, h;
};
// A Utility Function to check whether given cell (row, col)
// is a valid cell or not.
bool isValid(int row, int col)
{
// Returns true if row number and column number
// is in range
return (row >= 0) && (row < ROW) &&
(col >= 0) && (col < COL);
}
// A Utility Function to check whether the given cell is
// blocked or not
bool isUnBlocked(int grid[][COL], int row, int col)
{
// Returns true if the cell is not blocked else false
if (grid[row][col] == 1)
return (true);
else
return (false);
}
// A Utility Function to check whether destination cell has
// been reached or not
bool isDestination(int row, int col, Pair dest)
{
if (row == dest.first && col == dest.second)
return (true);
else
return (false);
}
// A Utility Function to calculate the 'h' heuristics.
double calculateHValue(int row, int col, Pair dest)
{
// Return using the distance formula
return ((double)sqrt ((row-dest.first)*(row-dest.first)
+ (col-dest.second)*(col-dest.second)));
}
// A Utility Function to trace the path from the source
// to destination
void tracePath(cell cellDetails[][COL], Pair dest)
{
printf ("\nThe Path is ");
int row = dest.first;
int col = dest.second;
stack<Pair> Path;
while (!(cellDetails[row][col].parent_i == row
&& cellDetails[row][col].parent_j == col ))
{
Path.push (make_pair (row, col));
int temp_row = cellDetails[row][col].parent_i;
int temp_col = cellDetails[row][col].parent_j;
row = temp_row;
col = temp_col;
}
Path.push (make_pair (row, col));
while (!Path.empty())
{
pair<int,int> p = Path.top();
Path.pop();
printf("-> (%d,%d) ",p.first,p.second);
}
return;
}
// A Function to find the shortest path between
// a given source cell to a destination cell according
// to A* Search Algorithm
void aStarSearch(int grid[][COL], Pair src, Pair dest)
{
// If the source is out of range
if (isValid (src.first, src.second) == false)
{
printf ("Source is invalid\n");
return;
}
// If the destination is out of range
if (isValid (dest.first, dest.second) == false)
{
printf ("Destination is invalid\n");
return;
}
// Either the source or the destination is blocked
if (isUnBlocked(grid, src.first, src.second) == false ||
isUnBlocked(grid, dest.first, dest.second) == false)
{
printf ("Source or the destination is blocked\n");
return;
}
// If the destination cell is the same as source cell
if (isDestination(src.first, src.second, dest) == true)
{
printf ("We are already at the destination\n");
return;
}
// Create a closed list and initialise it to false which means
// that no cell has been included yet
// This closed list is implemented as a boolean 2D array
bool closedList[ROW][COL];
memset(closedList, false, sizeof (closedList));
// Declare a 2D array of structure to hold the details
//of that cell
cell cellDetails[ROW][COL];
int i, j;
for (i=0; i<ROW; i++)
{
for (j=0; j<COL; j++)
{
cellDetails[i][j].f = FLT_MAX;
cellDetails[i][j].g = FLT_MAX;
cellDetails[i][j].h = FLT_MAX;
cellDetails[i][j].parent_i = -1;
cellDetails[i][j].parent_j = -1;
}
}
// Initialising the parameters of the starting node
i = src.first, j = src.second;
cellDetails[i][j].f = 0.0;
cellDetails[i][j].g = 0.0;
cellDetails[i][j].h = 0.0;
cellDetails[i][j].parent_i = i;
cellDetails[i][j].parent_j = j;
/*
Create an open list having information as-
<f, <i, j>>
where f = g + h,
and i, j are the row and column index of that cell
Note that 0 <= i <= ROW-1 & 0 <= j <= COL-1
This open list is implenented as a set of pair of pair.*/
set<pPair> openList;
// Put the starting cell on the open list and set its
// 'f' as 0
openList.insert(make_pair (0.0, make_pair (i, j)));
// We set this boolean value as false as initially
// the destination is not reached.
bool foundDest = false;
while (!openList.empty())
{
pPair p = *openList.begin();
// Remove this vertex from the open list
openList.erase(openList.begin());
// Add this vertex to the open list
i = p.second.first;
j = p.second.second;
closedList[i][j] = true;
/*
Generating all the 8 successor of this cell
N.W N N.E
\ | /
\ | /
W----Cell----E
/ | \
/ | \
S.W S S.E
Cell-->Popped Cell (i, j)
N --> North (i-1, j)
S --> South (i+1, j)
E --> East (i, j+1)
W --> West (i, j-1)
N.E--> North-East (i-1, j+1)
N.W--> North-West (i-1, j-1)
S.E--> South-East (i+1, j+1)
S.W--> South-West (i+1, j-1)*/
// To store the 'g', 'h' and 'f' of the 8 successors
double gNew, hNew, fNew;
//----------- 1st Successor (North) ------------
// Only process this cell if this is a valid one
if (isValid(i-1, j) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i-1, j, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i-1][j].parent_i = i;
cellDetails[i-1][j].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i-1][j] == false &&
isUnBlocked(grid, i-1, j) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue (i-1, j, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i-1][j].f == FLT_MAX ||
cellDetails[i-1][j].f > fNew)
{
openList.insert( make_pair(fNew,
make_pair(i-1, j)));
// Update the details of this cell
cellDetails[i-1][j].f = fNew;
cellDetails[i-1][j].g = gNew;
cellDetails[i-1][j].h = hNew;
cellDetails[i-1][j].parent_i = i;
cellDetails[i-1][j].parent_j = j;
}
}
}
//----------- 2nd Successor (South) ------------
// Only process this cell if this is a valid one
if (isValid(i+1, j) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i+1, j, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i+1][j].parent_i = i;
cellDetails[i+1][j].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i+1][j] == false &&
isUnBlocked(grid, i+1, j) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue(i+1, j, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i+1][j].f == FLT_MAX ||
cellDetails[i+1][j].f > fNew)
{
openList.insert( make_pair (fNew, make_pair (i+1, j)));
// Update the details of this cell
cellDetails[i+1][j].f = fNew;
cellDetails[i+1][j].g = gNew;
cellDetails[i+1][j].h = hNew;
cellDetails[i+1][j].parent_i = i;
cellDetails[i+1][j].parent_j = j;
}
}
}
//----------- 3rd Successor (East) ------------
// Only process this cell if this is a valid one
if (isValid (i, j+1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i, j+1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i][j+1].parent_i = i;
cellDetails[i][j+1].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i][j+1] == false &&
isUnBlocked (grid, i, j+1) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue (i, j+1, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i][j+1].f == FLT_MAX ||
cellDetails[i][j+1].f > fNew)
{
openList.insert( make_pair(fNew,
make_pair (i, j+1)));
// Update the details of this cell
cellDetails[i][j+1].f = fNew;
cellDetails[i][j+1].g = gNew;
cellDetails[i][j+1].h = hNew;
cellDetails[i][j+1].parent_i = i;
cellDetails[i][j+1].parent_j = j;
}
}
}
//----------- 4th Successor (West) ------------
// Only process this cell if this is a valid one
if (isValid(i, j-1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i, j-1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i][j-1].parent_i = i;
cellDetails[i][j-1].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i][j-1] == false &&
isUnBlocked(grid, i, j-1) == true)
{
gNew = cellDetails[i][j].g + 1.0;
hNew = calculateHValue(i, j-1, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i][j-1].f == FLT_MAX ||
cellDetails[i][j-1].f > fNew)
{
openList.insert( make_pair (fNew,
make_pair (i, j-1)));
// Update the details of this cell
cellDetails[i][j-1].f = fNew;
cellDetails[i][j-1].g = gNew;
cellDetails[i][j-1].h = hNew;
cellDetails[i][j-1].parent_i = i;
cellDetails[i][j-1].parent_j = j;
}
}
}
//----------- 5th Successor (North-East) ------------
// Only process this cell if this is a valid one
if (isValid(i-1, j+1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i-1, j+1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i-1][j+1].parent_i = i;
cellDetails[i-1][j+1].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i-1][j+1] == false &&
isUnBlocked(grid, i-1, j+1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i-1, j+1, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i-1][j+1].f == FLT_MAX ||
cellDetails[i-1][j+1].f > fNew)
{
openList.insert( make_pair (fNew,
make_pair(i-1, j+1)));
// Update the details of this cell
cellDetails[i-1][j+1].f = fNew;
cellDetails[i-1][j+1].g = gNew;
cellDetails[i-1][j+1].h = hNew;
cellDetails[i-1][j+1].parent_i = i;
cellDetails[i-1][j+1].parent_j = j;
}
}
}
//----------- 6th Successor (North-West) ------------
// Only process this cell if this is a valid one
if (isValid (i-1, j-1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination (i-1, j-1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i-1][j-1].parent_i = i;
cellDetails[i-1][j-1].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i-1][j-1] == false &&
isUnBlocked(grid, i-1, j-1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i-1, j-1, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i-1][j-1].f == FLT_MAX ||
cellDetails[i-1][j-1].f > fNew)
{
openList.insert( make_pair (fNew, make_pair (i-1, j-1)));
// Update the details of this cell
cellDetails[i-1][j-1].f = fNew;
cellDetails[i-1][j-1].g = gNew;
cellDetails[i-1][j-1].h = hNew;
cellDetails[i-1][j-1].parent_i = i;
cellDetails[i-1][j-1].parent_j = j;
}
}
}
//----------- 7th Successor (South-East) ------------
// Only process this cell if this is a valid one
if (isValid(i+1, j+1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i+1, j+1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i+1][j+1].parent_i = i;
cellDetails[i+1][j+1].parent_j = j;
printf ("The destination cell is found\n");
tracePath (cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i+1][j+1] == false &&
isUnBlocked(grid, i+1, j+1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i+1, j+1, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i+1][j+1].f == FLT_MAX ||
cellDetails[i+1][j+1].f > fNew)
{
openList.insert(make_pair(fNew,
make_pair (i+1, j+1)));
// Update the details of this cell
cellDetails[i+1][j+1].f = fNew;
cellDetails[i+1][j+1].g = gNew;
cellDetails[i+1][j+1].h = hNew;
cellDetails[i+1][j+1].parent_i = i;
cellDetails[i+1][j+1].parent_j = j;
}
}
}
//----------- 8th Successor (South-West) ------------
// Only process this cell if this is a valid one
if (isValid (i+1, j-1) == true)
{
// If the destination cell is the same as the
// current successor
if (isDestination(i+1, j-1, dest) == true)
{
// Set the Parent of the destination cell
cellDetails[i+1][j-1].parent_i = i;
cellDetails[i+1][j-1].parent_j = j;
printf("The destination cell is found\n");
tracePath(cellDetails, dest);
foundDest = true;
return;
}
// If the successor is already on the closed
// list or if it is blocked, then ignore it.
// Else do the following
else if (closedList[i+1][j-1] == false &&
isUnBlocked(grid, i+1, j-1) == true)
{
gNew = cellDetails[i][j].g + 1.414;
hNew = calculateHValue(i+1, j-1, dest);
fNew = gNew + hNew;
// If it isn't on the open list, add it to
// the open list. Make the current square
// the parent of this square. Record the
// f, g, and h costs of the square cell
// OR
// If it is on the open list already, check
// to see if this path to that square is better,
// using 'f' cost as the measure.
if (cellDetails[i+1][j-1].f == FLT_MAX ||
cellDetails[i+1][j-1].f > fNew)
{
openList.insert(make_pair(fNew,
make_pair(i+1, j-1)));
// Update the details of this cell
cellDetails[i+1][j-1].f = fNew;
cellDetails[i+1][j-1].g = gNew;
cellDetails[i+1][j-1].h = hNew;
cellDetails[i+1][j-1].parent_i = i;
cellDetails[i+1][j-1].parent_j = j;
}
}
}
}
// When the destination cell is not found and the open
// list is empty, then we conclude that we failed to
// reach the destiantion cell. This may happen when the
// there is no way to destination cell (due to blockages)
if (foundDest == false)
printf("Failed to find the Destination Cell\n");
return;
}
// Driver program to test above function
int main()
{
/* Description of the Grid-
1--> The cell is not blocked
0--> The cell is blocked */
int grid[ROW][COL] =
{
{ 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
{ 1, 1, 1, 0, 1, 1, 1, 0, 1, 1 },
{ 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 0, 0, 0, 0, 1 },
{ 1, 1, 1, 0, 1, 1, 1, 0, 1, 0 },
{ 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 },
{ 1, 0, 0, 0, 0, 1, 0, 0, 0, 1 },
{ 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
{ 1, 1, 1, 0, 0, 0, 1, 0, 0, 1 }
};
// Source is the left-most bottom-most corner
Pair src = make_pair(8, 0);
// Destination is the left-most top-most corner
Pair dest = make_pair(0, 0);
aStarSearch(grid, src, dest);
return(0);
}
you can try A* algo made for 9 row and 10 cols update it as per your requirement
您可以尝试为 9 行和 10 列制作的 A* 算法根据您的要求更新它
回答by Firer
Lets assume the ground as:
让我们假设地面为：
1 1 1 1 1 E 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 C 1 1 1 1 1
1 1 1 1 1 E 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 C 1 1 1 1 1
But always try to let the program know about properties of shapes e.g. square is equal on all sides so that's why a perfect diagonal way is shortest so if there are walls u could let your program choose the nearest part of the diagonal as
但是总是尽量让程序知道形状的属性，例如正方形在所有边上都是相等的，这就是为什么完美的对角线是最短的，所以如果有墙，你可以让你的程序选择对角线的最近部分作为
1 1 1 1 1 E 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 C 1 1 1 1 1
1 1 1 1 1 E 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 C 1 1 1 1 1
the part next to C (1) and then again continuing diagonally will help. Apologize for any grammatical or spelling mistakes.
C (1) 旁边的部分然后再次继续对角线将有所帮助。为任何语法或拼写错误道歉。
