如果（open_nodes_map[xdx][ydy]>；m0->；getPriority（）），这一行的其他含义是什么？ 我试图理解A*算法C++代码，下面的代码，我面临着一些难以理解的这一行。它在注释中说[[//已在开放列表中的节点：xdx，ydy，如果它们在当前点计算的优先级较高（较小的值），//则覆盖以前的优先级值并更改它们的父节点。]]。这里是指价值还是优先级 这是在A*算法C++代码之下 // Astar.cpp // http://en.wikipedia.org/wiki/A* // Compiler: Dev-C++ 4.9.9.2 // FB - 201012256 #include <iostream> #include <iomanip> #include <queue> #include <string> #include <math.h> #include <ctime> using namespace std; const int n=60; // horizontal size of the map const int m=60; // vertical size size of the map static int map[n][m]; static int closed_nodes_map[n][m]; // map of closed (tried-out) nodes static int open_nodes_map[n][m]; // map of open (not-yet-tried) nodes static int dir_map[n][m]; // map of directions const int dir=8; // number of possible directions to go at any position // if dir==4 //static int dx[dir]={1, 0, -1, 0}; //static int dy[dir]={0, 1, 0, -1}; // if dir==8 static int dx[dir]={1, 1, 0, -1, -1, -1, 0, 1}; static int dy[dir]={0, 1, 1, 1, 0, -1, -1, -1}; class node { // current position int xPos; int yPos; // total distance already travelled to reach the node int level; // priority=level+remaining distance estimate int priority; // smaller: higher priority public: node(int xp, int yp, int d, int p) {xPos=xp; yPos=yp; level=d; priority=p;} int getxPos() const {return xPos;} int getyPos() const {return yPos;} int getLevel() const {return level;} int getPriority() const {return priority;} void updatePriority(const int & xDest, const int & yDest) { priority=level+estimate(xDest, yDest)*10; //A* } // give better priority to going strait instead of diagonally void nextLevel(const int & i) // i: direction { level+=(dir==8?(i%2==0?10:14):10); } // Estimation function for the remaining distance to the goal. const int & estimate(const int & xDest, const int & yDest) const { static int xd, yd, d; xd=xDest-xPos; yd=yDest-yPos; // Euclidian Distance d=static_cast<int>(sqrt(xd*xd+yd*yd)); // Manhattan distance //d=abs(xd)+abs(yd); // Chebyshev distance //d=max(abs(xd), abs(yd)); return(d); } }; // Determine priority (in the priority queue) bool operator<(const node & a, const node & b) { return a.getPriority() > b.getPriority(); } // A-star algorithm. // The route returned is a string of direction digits. string pathFind( const int & xStart, const int & yStart, const int & xFinish, const int & yFinish ) { static priority_queue<node> pq[2]; // list of open (not-yet-tried) nodes static int pqi; // pq index static node* n0; static node* m0; static int i, j, x, y, xdx, ydy; static char c; pqi=0; // reset the node maps for(y=0;y<m;y++) { for(x=0;x<n;x++) { closed_nodes_map[x][y]=0; open_nodes_map[x][y]=0; } } // create the start node and push into list of open nodes n0=new node(xStart, yStart, 0, 0); n0->updatePriority(xFinish, yFinish); pq[pqi].push(*n0); open_nodes_map[x][y]=n0->getPriority(); // mark it on the open nodes map // A* search while(!pq[pqi].empty()) { // get the current node w/ the highest priority // from the list of open nodes n0=new node( pq[pqi].top().getxPos(), pq[pqi].top().getyPos(), pq[pqi].top().getLevel(), pq[pqi].top().getPriority()); x=n0->getxPos(); y=n0->getyPos(); pq[pqi].pop(); // remove the node from the open list open_nodes_map[x][y]=0; // mark it on the closed nodes map closed_nodes_map[x][y]=1; // quit searching when the goal state is reached //if((*n0).estimate(xFinish, yFinish) == 0) if(x==xFinish && y==yFinish) { // generate the path from finish to start // by following the directions string path=""; while(!(x==xStart && y==yStart)) { j=dir_map[x][y]; c='0'+(j+dir/2)%dir; path=c+path; x+=dx[j]; y+=dy[j]; } // garbage collection delete n0; // empty the leftover nodes while(!pq[pqi].empty()) pq[pqi].pop(); return path; } // generate moves (child nodes) in all possible directions for(i=0;i<dir;i++) { xdx=x+dx[i]; ydy=y+dy[i]; if(!(xdx<0 || xdx>n-1 || ydy<0 || ydy>m-1 || map[xdx][ydy]==1 || closed_nodes_map[xdx][ydy]==1)) { // generate a child node m0=new node( xdx, ydy, n0->getLevel(), n0->getPriority()); m0->nextLevel(i); m0->updatePriority(xFinish, yFinish); // if it is not in the open list then add into that if(open_nodes_map[xdx][ydy]==0) { open_nodes_map[xdx][ydy]=m0->getPriority(); pq[pqi].push(*m0); // mark its parent node direction dir_map[xdx][ydy]=(i+dir/2)%dir; } else if(open_nodes_map[xdx][ydy]>m0->getPriority()) { // update the priority info open_nodes_map[xdx][ydy]=m0->getPriority(); // update the parent direction info dir_map[xdx][ydy]=(i+dir/2)%dir; // replace the node // by emptying one pq to the other one // except the node to be replaced will be ignored // and the new node will be pushed in instead while(!(pq[pqi].top().getxPos()==xdx && pq[pqi].top().getyPos()==ydy)) { pq[1-pqi].push(pq[pqi].top()); pq[pqi].pop(); } pq[pqi].pop(); // remove the wanted node // empty the larger size pq to the smaller one if(pq[pqi].size()>pq[1-pqi].size()) pqi=1-pqi; while(!pq[pqi].empty()) { pq[1-pqi].push(pq[pqi].top()); pq[pqi].pop(); } pqi=1-pqi; pq[pqi].push(*m0); // add the better node instead } else delete m0; // garbage collection } } delete n0; // garbage collection } return ""; // no route found } int main() { srand(time(NULL)); // create empty map for(int y=0;y<m;y++) { for(int x=0;x<n;x++) map[x][y]=0; } // fillout the map matrix with a '+' pattern for(int x=n/8;x<n*7/8;x++) { map[x][m/2]=1; } for(int y=m/8;y<m*7/8;y++) { map[n/2][y]=1; } // randomly select start and finish locations int xA, yA, xB, yB; switch(rand()%8) { case 0: xA=0;yA=0;xB=n-1;yB=m-1; break; case 1: xA=0;yA=m-1;xB=n-1;yB=0; break; case 2: xA=n/2-1;yA=m/2-1;xB=n/2+1;yB=m/2+1; break; case 3: xA=n/2-1;yA=m/2+1;xB=n/2+1;yB=m/2-1; break; case 4: xA=n/2-1;yA=0;xB=n/2+1;yB=m-1; break; case 5: xA=n/2+1;yA=m-1;xB=n/2-1;yB=0; break; case 6: xA=0;yA=m/2-1;xB=n-1;yB=m/2+1; break; case 7: xA=n-1;yA=m/2+1;xB=0;yB=m/2-1; break; } cout<<"Map Size (X,Y): "<<n<<","<<m<<endl; cout<<"Start: "<<xA<<","<<yA<<endl; cout<<"Finish: "<<xB<<","<<yB<<endl; // get the route clock_t start = clock(); string route=pathFind(xA, yA, xB, yB); if(route=="") cout<<"An empty route generated!"<<endl; clock_t end = clock(); double time_elapsed = double(end - start); cout<<"Time to calculate the route (ms): "<<time_elapsed<<endl; cout<<"Route:"<<endl; cout<<route<<endl<<endl; // follow the route on the map and display it if(route.length()>0) { int j; char c; int x=xA; int y=yA; map[x][y]=2; for(int i=0;i<route.length();i++) { c =route.at(i); j=atoi(&c); x=x+dx[j]; y=y+dy[j]; map[x][y]=3; } map[x][y]=4; // display the map with the route for(int y=0;y<m;y++) { for(int x=0;x<n;x++) if(map[x][y]==0) cout<<"."; else if(map[x][y]==1) cout<<"O"; //obstacle else if(map[x][y]==2) cout<<"S"; //start else if(map[x][y]==3) cout<<"R"; //route else if(map[x][y]==4) cout<<"F"; //finish cout<<endl; } } getchar(); // wait for a (Enter) keypress return(0); }

根据本节的代码，您有以下问题：

// When the searching is over i.e. (x, y) is the final position, you try to
// recover the path from the start to the last position and return it as "path"

string path="";

while(!(x==xStart && y==yStart)) // we will be moving from final pos. to the start, backwards
{
  j=dir_map[x][y]; // direction map has the information about direction that you came from
  c='0'+(j+dir/2)%dir; // cast direction to single character...
  path=c+path; // ...to append it to a string path variable
  x+=dx[j]; // update x to previous x position (to recover path)
  y+=dy[j];  // update y to previous x position (to recover path)
}

换言之，您进行搜索（我假设您熟悉A*的工作原理），并为每个移动保存您选择的方向，因此基本上保存每个位置的路径

---

当你意识到你到达了最终位置后，你使用“方向图”来恢复你的后退（即从你到达实际位置（x，y）的位置）。执行从结尾到开头的步骤时，将连续的决策（方向）保存到字符串中

你能说得更具体些吗？“那你遇到麻烦的那一部分呢？”呜呜呜呜亲爱的先生；我提到了代码下面的部分：string path=“”；而（！（x==xStart&&y==yStart））{j=dir_-map[x][y]；c='0'+（j+dir/2）%dir；path=c+path；x+=dx[j]；y+=dy[j]；}这一个：我不明白。非常感谢亲爱的先生提供的信息，但是这里的“0”是什么意思？这个（j+dir/2）%dir。。。在这一行中，c='0'+（j+dir/2）%dir；我知道这项工作，但这个表达式是如何工作的？它是关于从整数到字符的转换

（j+dir/2）%dir

生成一个整数（一个数字），您希望将其保存在char-type变量中，以便以后可以在字符串中显示这几个数字。为此，您将

'0'

（或48-ASCII码）添加到该值中，并将一个数字作为可附加到字符串的字符。您好，亲爱的先生。在获得您的许可后，能否澄清从注释（//生成所有可能方向的移动（子节点））开始到（返回“”；//未找到路由）结束的部分。请

// When the searching is over i.e. (x, y) is the final position, you try to
// recover the path from the start to the last position and return it as "path"

string path="";

while(!(x==xStart && y==yStart)) // we will be moving from final pos. to the start, backwards
{
  j=dir_map[x][y]; // direction map has the information about direction that you came from
  c='0'+(j+dir/2)%dir; // cast direction to single character...
  path=c+path; // ...to append it to a string path variable
  x+=dx[j]; // update x to previous x position (to recover path)
  y+=dy[j];  // update y to previous x position (to recover path)
}