/* 云风求解最短路径代码 (Cloud Wu\'s Pathfinding code)
* 1999 年 1月 8 日 (1999, Jan 8)
* 这段代码没有进行任何优化(包括算法上), 但不意味我不知道该怎样优化它
* 它是为教学目而做,旨在用易于理解和简洁代码描述出 A* 算法在求最段路
* 径中运用. 由于很久没有摸算法书, 本不能保证是纯正 A* 算法 ;-)
* 你可以在理解了这段基础上,按自己理解写出类似代码. 但是简单
* 复制它到你中是不允许,如果你真要这样干,请在直接使用它软件Software
* 文档中,写上我名字 ;-)
* 有任何问题,或建议请 E-mail 到 [email protected]?
* 欢迎参观我主页 http://member.netease.com/~cloudwu (云风工作室)
* (你可以在上面找到些有关这个问题讨论,和有关游戏设计其它大量资料)
*
* 本附带有个数据文件 map.dat, 保存有地图数据
*/
// # NDEBUG
# <stdio.h>
# <conio.h>
# <assert.h>
# <stdlib.h>
# MAPMAXSIZE 100 //地图面积最大为 100x100
# MAXINT 8192 //定义个最大整数, 地图上任意两点距离不会超过它
# STACKSIZE 65536 //保存搜索节点堆栈大小
# tile_num(x,y) ((y)*map_w+(x)) //将 x,y 坐标转换为地图上块编号
# tile_x(n) ((n)%map_w) //由块编号得出 x,y 坐标
# tile_y(n) ((n)/map_w)
// 树结构, 比较特殊, 是从叶节点向根节点反向链接
typedef struct node *TREE;
struct node {
??? h;
??? tile;
??? TREE father;
} ;
typedef struct node2 *LINK;
struct node2 {
??? TREE node;
??? f;
??? LINK next;
};
LINK queue; // 保存没有处理行走思路方法节点
TREE stack[STACKSIZE]; // 保存已经处理过节点 (搜索完后释放)
stacktop;
unsigned char map[MAPMAXSIZE][MAPMAXSIZE]; //地图数据
dis_map[MAPMAXSIZE][MAPMAXSIZE]; //保存搜索路径时,中间目标地最优解
map_w,map_h; //地图宽和高
start_x,start_y,end_x,end_y; //地点,终点坐标
// 化队列
void init_queue
{
??? queue=(LINK)malloc((*queue));
??? queue->node=NULL;
??? queue->f=-1;
??? queue->next=(LINK)malloc((*queue));
??? queue->next->f=MAXINT;
??? queue->next->node=NULL;
??? queue->next->next=NULL;
}
// 待处理节点入队列, 依靠对目地估价距离插入排序
void enter_queue(TREE node, f)
{
??? LINK p=queue,father,q;
??? while(f>p->f) {
??????? father=p;
??????? p=p->next;
??????? assert(p);
??? }
??? q=(LINK)malloc((*q));
??? assert(queue);
??? q->f=f,q->node=node,q->next=p;
??? father->next=q;
}
// 将离目地估计最近方案出队列
TREE get_from_queue
{
??? TREE bestchoice=queue->next->node;
??? LINK next=queue->next->next;
??? free(queue->next);
??? queue->next=next;
??? stack[stacktop]=bestchoice;
??? assert(stacktop<STACKSIZE);
??? bestchoice;
}
// 释放栈顶节点
void pop_stack
{
??? free(stack[--stacktop]);
}
// 释放申请过所有节点
void freetree
{
??? i;
??? LINK p;
??? for (i=0;i<stacktop;i)
??????? free(stack[i]);
??? while (queue) {
??????? p=queue;
??????? free(p->node);
??????? queue=queue->next;
??????? free(p);
??? }
}
// 估价,估价 x,y 到目地距离,估计值必须保证比实际值小
judge( x, y)
{
??? distance;
??? distance=abs(end_x-x)+abs(end_y-y);
??? distance;
}
// 尝试下步移动到 x,y 可行否
trytile( x, y,TREE father)
{
??? TREE p=father;
??? h;
??? (map[y][x]!=\' \') 1; // 如果 (x,y) 处是障碍,失败
??? while (p) {
??????? (xtile_x(p->tile) && ytile_y(p->tile)) 1; //如果 (x,y) 曾经经过,失败
??????? p=p->father;
??? }
??? h=father->h+1;
??? (h>=dis_map[y][x]) 1; // 如果曾经有更好方案移动到 (x,y) 失败
??? dis_map[y][x]=h; // 记录这次到 (x,y) 距离为历史最佳距离
??? // 将这步方案记入待处理队列
??? p=(TREE)malloc((*p));
??? p->father=father;
??? p->h=father->h+1;
??? p->tile=tile_num(x,y);
??? enter_queue(p,p->h+judge(x,y));
??? 0;
}
// 路径寻找主
void findpath( *path)
{
??? TREE root;
??? i,j;
??? stacktop=0;
??? for (i=0;i<map_h;i)
??????? for (j=0;j<map_w;j)
??????????? dis_map[i][j]=MAXINT;
??? init_queue;
??? root=(TREE)malloc((*root));
??? root->tile=tile_num(start_x,start_y);
??? root->h=0;
??? root->father=NULL;
??? enter_queue(root,judge(start_x,start_y));
??? for (;;) {
??????? x,y,child;
??????? TREE p;
??????? root=get_from_queue;
??????? (rootNULL) {
??????????? *path=-1;
??????????? ;
??????? }
??????? x=tile_x(root->tile);
??????? y=tile_y(root->tile);
??????? (xend_x && yend_y) ; // 达到目地成功返回
??????? child=trytile(x,y-1,root); //尝试向上移动
??????? child&=trytile(x,y+1,root); //尝试向下移动
??????? child&=trytile(x-1,y,root); //尝试向左移动
??????? child&=trytile(x+1,y,root); //尝试向右移动
??????? (child!=0)
??????????? pop_stack; // 如果 4个方向均不能移动,释放这个死节点
??? }
??? // 回溯树将求出最佳路径保存在 path 中
??? for (i=0;root;i) {
??????? path[i]=root->tile;
??????? root=root->father;
??? }
??? path[i]=-1;
??? freetree;
}
void prpath( *path)
{
??? i;
??? for (i=0;path[i]>=0;i) {
??????? gotoxy(tile_x(path[i])+1,tile_y(path[i])+1);
??????? cprf(\"\\xfe\");
??? }
}
readmap
{
??? FILE *f;
??? i,j;
??? f=fopen(\"map.dat\",\"r\");
??? assert(f);
??? fscanf(f,\"%d,%d\\n\",&map_w,&map_h);
??? for (i=0;i<map_h;i)
??????? fgets(&map[i][0],map_w+1,f);
??? fclose(f);
??? start_x=-1,end_x=-1;
??? for (i=0;i<map_h;i)
??????? for (j=0;j<map_w;j) {
??????????? (map[i][j]\'s\') map[i][j]=\' \',start_x=j,start_y=i;
??????????? (map[i][j]\'e\') map[i][j]=\' \',end_x=j,end_y=i;
??????? }
??? assert(start_x>=0 && end_x>=0);
??? 0;
}
void showmap
{
??? i,j;
??? clrscr;
??? for (i=0;i<map_h;i) {
??????? gotoxy(1,i+1);
??????? for (j=0;j<map_w;j)
??????? (map[i][j]!=\' \') cprf(\"\\xdb\");
??????? cprf(\" \");
??? }
??? gotoxy(start_x+1,start_y+1);
??? cprf(\"s\");
??? gotoxy(end_x+1,end_y+1);
??? cprf(\"e\");
}
{
??? path[MAXINT];
??? readmap;
??? showmap;
??? getch;
??? findpath(path);
??? prpath(path);
??? getch;
??? 0;
}
运行另需要个描述地图数据文件 map.dat 如下
80,24oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo oo oo s oooooooooooooo oo o oooooooooooo o oo o ooooooo oooooooooooooo oooooooo oo oooooo o oooo o o oo o o ooo ooo oo oooo oooo oo ooooooooooooooooooooooooooooooooooooooooooooooo oo oo ooooooooooooooooooooooooooooooooooooooooooooo oo o oooooooooooo o ooooooo oooooooo oo o o o o oo ooooooooooo oooooooooo o o oo oe ooo o o oo ooooo o o o oo o o oo o o o oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
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